assignment based on topic — 1 . 1
TRANSCRIPT
(i) Law of conservation of mass : It states that mass can neither be created nor be destroyed ina chemical reaction, i.e., total mass of reactants must be equal to the total mass of products.
Law of conservation of mass is also known as “Law of indestructibility of matter.”(ii) Law of constant proportion : It states, that in a pure chemical substance, the elements are
always present in a definite proportion by mass.It is also known as law of definite proportions.
VERY SHORT ANSWER TYPE QUESTIONS
Q. 1. State the law of conservation of mass. [SA-II, H H H H H ]
Ans. Mass can neither be created nor be destroyed in a chemical reaction.
Q. 2. What is the law of constant proportions ? [CBSE, SA-II, H H H H ]
Ans. It states that a pure chemical compound is always found to be made-up of same elementscombined together in the same fixed ratio of mass.
Q. 3. Carbon dioxide collected from different sources contains carbon and oxygen in thesame proportion. Which law of chemical combination governs this ? [CBSE, H H H ]
Ans. This is governed by law of definite proportion.
Q. 4. Carbon dioxide produced by heating sodium hydrogen carbonate is dry whereas thatproduced by the action of dilute hydrochloric acid on sodium hydrogen carbonate ismoist. What do you think about the difference in the composition of carbon dioxidein the two cases ? [HOTS, SA-II, H H H H H ]
Ans. There will be no difference in the composition of CO2 in two cases, i.e., in both cases, Cand O will be present in the same ratio by mass.
Q. 5. A student performed an activity to verify the law of conservation of mass using solutionX and Y. Identify the solution ‘Y’ which can’t react with ‘X’ namely sodium chloride.
[SA-II, H H H H ]
Ans. Lead nitrate or silver nitrate
Laws of Chemical Combination
Assignment Based on Topic — 1 . 1
ATOMS AND MOLECULESCLASS–IX
— BY B.K. SINGH
1
2
SHORT ANSWER TYPE QUESTIONS [TYPE—I]Q. 1. (a) If 100 g of calcium carbonate on heating produces 44 g of carbon dioxide, how
much quick lime will be formed ?(b) Which law follows from this example ? Define the law. [HOTS, H H H H H ]
Ans. (a) Quick lime formed will be 56 g.(b) Law of conservation of mass follows from it because total mass of reactants
(100 g) = total mass of product (44 g + 56 g = 100 g).Q. 2. Hydrogen and oxygen combine in the ratio 1 : 8 by mass to form water. What mass ofoxygen
gas would be required to react completely with 3g of hydrogen gas ? [SA-II, H H H ]Ans. 1g of hydrogen reacts with oxygen = 8 g
3g of hydrogen reacts with oxygen = 8 × 3 = 24 gHence, mass of oxygen required is 24 g.
Q. 3. Potassium chlorate decomposes on heating to form potassium chloride and oxygen.When 24.5g of potassium chlorate is decomposed completely, 14.9g of potassium chlorideis formed. Calculate the mass of oxygen formed. [CBSE 2011, H H H ]
Ans. Potassium chlorate Heat Potassium chloride + Oxygen24.5g 14.9g x g
Let the mass of oxygen produced be x gThen x + 14.9 = 24.5 or x = 9.6 gThus, 9.6 g of oxygen is produced.
Q. 4. (a) What mass of silver nitrate will react with 58.5 g of sodium chloride to produce143.5 g of silver chloride and 85 g of sodium nitrate ?
(b) On what law is the above reaction based and state the law. (Atomic mass Ag = 108,N = 14, O = 16, Na = 23 and Cl = 35.5) [SA-II, H H H ]
Ans. (a) AgNO3 + NaCl AgCl + NaNO3We know that, mass of reactants = mass of products
5.85 + mass of silver nitrate = 14.35 + 8.55.85 + mass of silver nitrate = 22.85
mass of silver nitrate = 22.85 – 5.85 = 17 g(b) The above reaction is based on the law of constant proportions.Law of constant proportions states that “in a chemical substances, elements are alwayspresent in a definite proportion by weight.”
Q. 5. When 3.0 g of carbon is burnt in 8.0 g oxygen, 11.0 g of carbondioxide will be formed.When 3.0 g of carbon is burnt in 50.0 g of oxygen. Which law of chemical combinationwill govern your answer ? [H H H ]
Ans. The reaction of burning of carbon in oxygen may be written as :
2 2Carbon Oxygen Carbon dioxide1 mole 1 mole 1 mole
12 g 2 16 = 32 g 12 2 16 = 44 g
C O CO
It shows that 12 g (1 mole) of carbon burns in 32 g (1 mole) of oxygen to form 44 g (1 mole)of carbon dioxide. Therefore, 3 g of carbon reacts with 8 g of oxygen to form 11 g ofcarbon dioxide.It is given that, 3.0 g of carbon is burnt in 8.0 of oxygen to produce 11.0 g of carbon dioxide.Consequently 11.0 g of carbon dioxide will be formed when 3.0 g of carbon is burnt in50.0 g of oxygen consuming 8 g of oxygen, leaving behind 50 – 8 = 42.0 of oxygen.This answer governs the law of constant or definite proportion.
3
SHORT ANSWER TYPE QUESTIONS [TYPE—II]Q. 1. In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products
were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium ethanoate. Show that theseobservations are in agreement with the law of conservation of mass.Sodium carbonate + ethanoic acid sodium ethanoate + carbon dioxide + water
[H H H H ]Ans. Na2CO3 + 2CH3COOH 2CH3COONa + CO2 + H2O
5.3g 6g 8.2 xg 0.9g
Mass of reactants = Mass of Sodium Carbonate + Mass of Ethanoic acid= 5.3g + 6g = 11.3g
Mass of products = Mass of Sodium ethanoate + Mass of Carbon dioxide+ Mass of water
= 8.2 g + 2.2 g + 0.9 g = 11.3 Mass of reactants = Mass of products.Hence, these observation are in agreement with law of conservation of mass.
Q. 2. A sample of ammonia contains 9 g hydrogen and 42 g nitrogen. Another sample contains5 g hydrogen, calculate the amount of nitrogen in the second sample.
[H H H H H ]Ans. The ratio of hydrogen and nitrogen in the first sample of ammonia is 9 : 42 or 3 : 14
According to “law of definite proportions” the second sample of ammonia should alsocontain hydrogen and nitrogen in the ratio 3 : 14.
If hydrogen is 5 g, nitrogen is5 14
3
= 23.3 g.
Q. 3. Copper oxide was prepared by two different methods. In one case, 1.75 g of the metalgave 2.19 g of oxide. In the second case, 1.14 g of the metal gave 1.43 g of the oxide. Showthat the given data illustrate the law of constant proportions. [HOTS, H H H H H ]
Ans. In Case I : Mass of copper = 1.75 gMass of copper oxide = 2.19 g
% of copper in the oxide =Mass of copper
100Mass of copper oxide
=1.75
1002.19 = 79.9%
% of oxygen = 100 – 79.9 = 20.1%In Case II : Mass of copper = 1.14 g
Mass of copper oxide = 1.43 g
% of copper in the oxide =1.14
1001.43 = 79.7%
% of oxygen = 100 – 79.7 = 20.3%Thus, copper oxide prepared by any of the given methods contains copper and oxygen inthe same proportion by mass. Hence, it proves the law of constant proportions.
4
LONG ANSWER TYPE QUESTIONSQ. 1. Give an activity to prove that there is no change in mass during a chemical reaction.
[H H H ]Ans. (i) Take 10 mL of sodium chloride solution in the conical flask.
(ii) Take 10 mL of the silver nitrate solution in atest tube.
(iii) Suspend the test tube into the flask with thehelp of a thread carefully.
(iv) Close the mouth of the flask with a cork.(v) Weigh the flask along with its contents.(vi) Now, list the test tube and allow the two
solution to get mixed. Let the reaction takeplace. A white precipitate of silver chloridewill be formed.
(vii)Weigh the flask along with its contents.Observation : Mass of the reactants is equal to the mass of products.Conclusion : Law of conservation of mass is proved experimentally.
Q. 2. With the help of law of conservation of mass, predict the values of x, y and z in thefollowing reaction :
A + B C + D(i) 1.0 g 2.5 g 1.5 g x(ii) y 5.0 g 3.0 g 4.0 g(iii) 1.5 3.8 g z 3.0 g [HOTS, H H H H ]
Ans. According to law of conservation of mass, the total mass of reactants is equal to the totalmass of the products.(i) Mass of reactants = 1.0 + 2.5 = 3.5 g
Mass of products = 1.5 + x = (1.5 + x) g 3.5 g = 1.5 + x
x = 3.5 – 1.5 = 2.0 g(ii) Mass of reactants = y + 5.0
Mass of products = 3.0 + 4.0 = 7.0 y + 5.0 = 7.0or y = 7.0 – 5.0 = 2.0 g(iii) Mass of reactants = 1.5 + 3.8 = 5.3g
Mass of products = z + 3.0 5.3g = z + 3.0or z = 5.3 – 3.0 = 2.3 g
Atoms : It is the smallest particle of an element that maintains its chemical identity through outall chemical and physical changes.
CorkThread
Conical flask
Small ignition tubeSolution of AgNO3
Solution of NaCI
Fig. 1.1. To verify the law ofconservation of mass.
Atoms and Atomic Theory of Matter and Atomic Symbols
Assignment Based on Topic — 1 . 2
5
Daltons’s Atomic theory(i) Every matter is made of very small particles called the atoms.
(ii) Atoms are indivisible particles which can not be careated nor be destroyed in a chemicalreaction.
(iii) Atoms of given element are identical in mass as well as in chemical properties.(iv) Atoms of different elements have different masses and chemical properties.(v) Atoms combine in the ratio or small whole numbers forming compounds.
(vi) The relative number as well as kinds of atoms are constant in a given compound.Draw backs of Dalton’s Atomic Theory : (i) According to the modern theory, atom is not the
ultimate indivisible particle of matter. Atom are divisible, i.e., they are themselves made-up ofparticles (protons, electrons, neutrons etc.).
(ii) In case of isotopes of an element, the assumption that the atoms of the same element havesame mass does not hold good.
Atomic Symbols : A symbol signifies a shorthand representation of an atom of an element. Thesymbol of any element is based on the English name or Latin name and is represented by using onlyits first letter or the first letter and another letter.
(i) The first letter is written in capital and the second letter in small.(ii) The second letter can be any letter in the name of the element, which is internationally
accepted.Symbols for Some Elements
Element Symbol Element Symbol Element Symbol
Aluminium Al Copper Cu Nitrogen NArgon Ar Fluorine F Oxygen OBarium Ba Gold Au Potassium KBoron B Hydrogen H Silicon Si
Bromine Br Iodine I Silver AgCalcium Ca Iron Fe Sodium NaCarbon C Lead Pb Sulphur S
Chlorine Cl Magnesium Mg Uranium UCobalt Co Marcury Hg Zinc Zn.
Chromium Cr Neon Ne Artatine At
Elements named after places : Scandium (Sc), found in earlier name of Scandinavia, Europium(Eu) – after the continent Europe, Polonium (Po) – named after the curies after their home town inPoland.
Named after Planets : Selenium (Se) – ‘Seles’ Greek name for the Moon, Plutonium (Pu) –Neptunium (Np), Uranium (U), Mercury (Hg) was named after ‘a planet but derives its symbol Hgfrom the Latin worde Hydragyrum’ meaning liquid silver.
Named after Scientists : Curium (Cm) after Pierre and Marie curie, Fermium (Fm) after EnricoFermi, Einsteinium (Es) after Albert Einstein, Mendelevium (Md) after Dimition Mendeleev.
VERY SHORT ANSWER TYPE QUESTIONSQ. 1. What is an atom ? [SA-II, H H H H H ]Ans. An atom is defined as the smallest particle of an element which may or may not be capable of free
existence. However, it is the smallest particle that takes part in a chemical reaction.
6
Q. 2. Why is it not possible to see an atom with naked eyes ? [H H H H ]Ans. An atom is an extremely small particle whose radius is of the order to 10–10m. This size
is so small that our eyes are not able to see it.Q. 3. Name the smallest particle that has all the properties of an element. [H H H ]Ans. Atom.Q. 4. State the postulate of Dalton’s atomic theory which can explain the law of conservation
of mass. [CBSE 2011, SA-II, H H H H H ]Ans. According to a postulate of Dalton’s theory, atoms are indivisible particles which can not
be created or destroyed in a chemical reaction. This postiulates explains the law ofconservation of mass.
Q. 5. How do atoms exist ? [H H H ]Ans. Atoms exist in the form of atom, molecule or ions.Q. 6. Why is the symbol of silver written as Ag ? [HOTS, H H H ]Ans. The symbol of silver is derived from the Latin name, ‘Argentum’ from which the symbol
Ag has been derived.
SHORT ANSWER TYPE QUESTIONS [TYPE—I]Q. 1. (a) What was the postulate of Dalton’s atomic theory about the masses of atoms ?
(b) How was the postulate modified ? Give one example. [HOTS, H H H H ]
Ans. (a) According to Dalton’s atomic theory, atom of an element have same mass.
(b) The modified postulate states that atoms of the same element may have differentmasses, e.g., atoms of chlorine have atomic masses 35 and 37.
Q. 2. Which of the following symbols of elements are incorrect ? Give their correct symbol.
(a) Cobalt CO, (b) Aluminium, AL
(c) Helium, He (d) Sodium SO [NCERT Exemplar]
Ans. (a) Cobalt = CO i.e., 2nd letter should be small.
(b) Aluminium = Al, i.e., 2nd letter should be small
(c) Helium = He is correct.
(d) Sodium = Na (from latin name, Natrium)
Q. 3. Give the unit to measure size of atom and give size of hydrogen atom. [H H H H ]
Ans. The unit to measure size of atom is nano meter. Size of hydrogen atom is 10–10 m.
Q. 4. Give the Latin name for sodium, potassium, gold and mercury. [H H H ]
Ans. Sodium – Natrium
Gold – Aurum
Potassium – Kalium
Mercury – Hydragyrum.
Q. 5. Why was the Dalton’s atomic theory modified ? [SA-II, H H H H H ]
Ans. Nothing about the structure of atom was known at the time Dalton proposed his theory.
Due to discovery of electrons, protons, neutrons and isotopes, a need of certainmodification in Dalton’s atomic theory was felt.
7
SHORT ANSWER TYPE QUESTIONS [TYPE—II]Q. 1. State any three drawbacks of Dalton’s atomic theory of matter. [SA-II, H H H H H ]Ans. Drawbacks of Dalton’s atomic theory of matter are :
(i) According to Dalton’s atomic theory, matter is indivisible, i.e., which can not bedivided. But now it is known that under special circumstances, atoms can be dividedinto still smaller particles called electrons, protons and neutrons.
(ii) Dalton’s atomic theory fails to explain why substances like charcoal, graphite anddiamond have different properties when all these substances are made up of the sametype of atoms, called carbon atoms.
(iii) Dalton’s atomic theory postulated that all the atoms of the same element have exactly thesame mass. It is now known that isotopes are atoms of the element but have differentmasses.
Q. 2. What is the significance of the symbol of an element ? [HOTS, H H H H ]Ans. (i) The symbol stands for the name of the element.
(ii) The symbol stands for one atom of the element.(iii) The symbol represents quantity of the element equal in mass to its atomic mass, or
gram-atomic mass.(iv) The symbol also represents mass of the element which contains one Avogadro’s
number of atoms of that element.Q. 3. What is the qualitative meaning mass of the symbol of chlorine (Cl) of atomic mass
35.5u ? [CBSE 2010]Ans. (a) The symbol Cl represents that one atom of chlorine is 35.5 times heavier than one
atomic mass unit.(b) The symbol Cl represents that its atomic mass is 35.5 u.(c) The symbol Cl represents that one mole of Cl (35.5 g) will have 6.022 × 1023 atoms.
LONG ANSWER TYPE QUESTIONSQ. 1. What are the postulates of Dalton’s atomic theory of matter ? [SA-II, H H H H H ]Ans. Postulates of Dalton’s atomic theory of matter are :
(i) Each substance is made’ up of extremely small particles called atoms.(ii) Atoms of a given element are identical in mass and chemical properties.(iii) Elements are substances having the same kind of atoms. In other words, atoms of a
particular element are all alike but differ from atoms of other elements.(iv) Atoms can not be created nor destroyed.(v) Atoms of elements combine in the ratio of whole numbers to produce a large number
of compound-atoms of a new substance. The compound atoms of a particularsubstance are identical in all properties, and differ from that of other substances.
Q. 2. How will you explain the laws of constant proportion on the basis of Dalton’s atomictheory ? [HOTS, SA-II, H H H H ]
Ans. According to Dalton’s atomic theory, each element consists of atoms which are similarand have same weights. Further atoms of one element combine with atoms of anotherelement to form compounds. Let us suppose that x atoms of element A combine with y
8
atoms of element B and the compound formed is AxBy. If ‘a’ stands for the atomic massof A and ‘b’ for that of B, there
Percentage of A in the compound =100ax
ax by
Percentage of B in the compound =100by
ax by
Now a and b (atomic masses of elements) are fixed, x and y are also fixed whole numbersaccording to atomic theory. Therefore, precent of A and B in the compound is also constant.This shows that the composition of various elements in a compound is also fixed. This isthe law of constant proportion.
Molecules of Elements : The molecules of an element are contributed by the same type ofatoms. Molecules of many elements are made up of one atom of that element, e.g., argon (Ar), helium(He) etc.
The molecules of the most of the non-metals are made up of more than one atoms.For example, a molecule of oxygen consists of two atoms of oxygen. Ozone consists of three
atoms of oxygen.Atomicity : The number of atoms present in a molecule of an element is called its atomicity.
Atomicity of Some elements
Type of element Name Atomicity
Non-metal Argon MonoatomicHelium MonoatomicOxygen DiatomicHydrogen DiatomicNitrogen DiatomicChlorine DiatomicPhosphorus Tetra-atomicSulphur Poly-atomic
Metal Sodium MonoatomicIron MonoatomicAluminium MonoatomicCopper Monoatomic
Molecules of Compounds : Atoms of different elements join together in definite proportionsto form molecules of compounds.
Molecule : A molecule is smallest particle of an element or compound, which iscapable of independent existence.
Assignment Based on Topic — 1 . 3
9
Molecules of Some Compounds
Compound Combining elements Ratio by mass
Water Hydrogen, Oxygen 1 : 8Ammonia Nitrogen, Hydrogen 14 : 3Carbondioxide Carbon, Oxygen 3 : 8Glucose Carbon, hydrogen, Oxygen 6 : 1 : 8Common salt Sodium, Chlorine 23 : 3.5
VERY SHORT ANSWER TYPE QUESTIONSQ. 1. What is a molecule ? [H H H ]
Ans. A molecule is the smallest particle of an element or a compound which can exist freelyunder ordinary conditions and shows all the properties of that substance.
Q. 2. What do you mean by atomicity of a molecule ? [H H H H ]
Ans. The number of atoms present in one molecule of substance is called its atomicity.
Q. 3. Give two examples of polyatomic molecules of elements. [CBSE 2014, H H H H ]
Ans. (i) Phosphorus (P4), (ii) Sulphur (S8) .
Q. 4. What is the difference between an atom and a molecule ? [HOTS, SA-II, H H H H H ]
Ans. Atom is the smallest particle of an element which may or may not be capable of freeexistence where as molecule is the smallest particle of an element or a compound whichcan exist freely.
Q. 5. Give an example of a triatomic molecule of an element. [CBSE 2010, H H H ]
Ans. Ozone (O3).
SHORT ANSWER TYPE QUESTIONS [TYPE—I]
Q. 1. Classify the following on the basis of their atomicity : [H H H H ](i) Chlorine, (ii) phosphorus, (iii) helium, (iv) ozone.
Ans. Molecules (of elements) Atomicity
(i) Chlorine (Cl2) 2(ii) Phosphorus (P4) 4
(iii) Helium (He) 1(iv) Ozone (O3) 3
Q. 2. State the number of atoms present in each of the following chemical species :[HOTS, SA-II, H H H H H ]
(a) 2–3
CO (b) 3–4
PO
(c) P2O5 (d) CO
Ans. (a) 2–3
CO = 4 atoms (1 atom of carbon + 3 atoms of oxygen)
10
(b) 3–4
PO = 5 atoms (1 atom of phosphorus + 4 atoms of oxygen)
(c) P2O5 = 7 atoms (2 atoms of carbon + 5 atom of oxygen)(d) CO = 2 atoms (1 atom of carbon + 1 atom of oxygen)
Q. 3. Give the names of the elements present in the following compounds :(a) Quick lime (b) Hydrogen bromide(c) Baking powder (d) Potassium Sulphate [H H H H ]
Ans. (a) Quick lime [CaO] Elements (calcium and oxygen)(b) Hydrogen bromide (HBr) Element (Hydrogen + bromine)(c) Baking powder (NaHCO3) Element (Sodium, hydrogen, carbon and oxygen)(d) Potassium sulphate (K2SO4) Elements (Potassium, Sulphur and Oxygen)
Q. 4. Write the cations and anions present (if any) in the following compounds :(a) CH3COONa (b) NaCl(c) H2 (d) NH4NO3
[NCERT Exemplar, SA-II, H H H H ]
Ans. Compound Cation Anion
(a) CH3 COONa Na+ CH3 COO–
(b) NaCl Na+ Cl–
(c) H2 – –(Covalent compound, no ions)
(d) NH4NO3+4NH –
3NO
Q. 5. How would you differentiate between a molecule of an element and a molecule of acompound ? Write one example of each type. [SA-II, H H H H H ]
Ans. Molecule of an element is made up only one kind of atoms for example, Cl2, O2, S8.Molecule of a compound is made up of two or more different kinds of atoms in a fixedratio. For example, NaCl, CuSO4, H2O.
SHORT ANSWER TYPE QUESTIONS [TYPE—II]Q. 1. (a) What do the following formula stand for :
(i) 2 O, (ii) O2 (iii) O3 (iv) H2O(b) Define poly atomic ions. Write example. [CBSE 2013, 2015, H H H H H ](c) Separate the following stable elements into atoms and molecules.
Oxygen, Nitrogen, Argon, Sodium, Neon, Chlorine [CBSE 2011]Ans. (a) (i) Two atoms of oxygen.
(ii) Diatomic oxygen O2 molecule(iii) Triatomic oxygen O3 molecule(iv) Two atoms of hydrogen and 1 atom of oxygen forming one molecule of water.
(b) Cluster of atoms that act as an ion are called polyatomic ions.
For example : +4NO , 3–
4PO
(c) Atoms : Argon, Sodium, NeonMolecules : Oxygen, Nitrogen, Chlorine
11
Q. 2. Comment on the statement, “A molecule is the smallest particle of a substance (elementor compound) which has the properties of the substance and is stable.”
[HOTS, SA-II, H H H ]Ans. Yes, it is right that a molecule is the smallest particle of a substance which has the
properties, of that substance and is stable. In case of gases like oxygen, a molecule (O2) isstable and gives properties of oxygen gas, whereas atom ‘O’ is not stable and does notrepresent oxygen gas. In case of noble gases such as helium, a molecule is of monoatomichelium atom (He) and represents helium gas and is also stable. In case of compound, forexample, carbon dioxide, properties of carbon dioxide molecule are the properties of carbondioxide gas. A molecule of carbon dioxide is stable.
Q. 3. What are ionic and molecular compounds ? Give examples. [NCERT Exemplar, H H H ]Ans. Ionic compounds : The compounds in which the constituent particles are ions. Examples
: Sodium Chloride, Calcium oxide are ionic compounds.Molecular compounds : The compounds in which the constituent particle are discretemolecule are called molecular compounds. Examples : Water, Ammonia, Carbon dioxideare molecular compounds.
LONG ANSWER TYPE QUESTIONS
Q. 1. (a) Atoms of inert gas elements are mono-atomic while that of the other elements arenot. Assign reason.
(b) Define atomicity of an element and give one example each of mono, di and triatomicmolecules. [SA-II, H H H H H ]
Ans. (a) The atoms of inert gas elements have stable electronic configurations. Therefore, theycan exist independently and are monoatomic in nature. The atoms of all other elementsare yet to have stable configuration. To achieve it, they combine with the atoms of theother elements.
(b) The number of atoms combined together to from a stable molecule of an element iscalled its atomicity.
Examples :Monoatomic molecule : helium (He), neon (Ne)Diatomic molecule : hydrogen (H2), oxygen (O2)Triatomic molecule : Ozone (O3), Carbondioxide (CO2)Tetra atomic molecule : P4Polyatomic molecule : S8
Q. 2. (a) What is the difference between :(i) an atom of hydrogen and a molecule of hydrogen.
(ii) a molecule of oxygen and a molecule of water ?(b) What is an ion ? Write the symbols of the following :
Sodium ion, oxide ion, polyatomic anion, polyatomic cation. [CBSE 2013, H H H H ]Ans. (a) (i) An atom of hydrogen is the smallest particle of hydrogen element which does not
have independent existence. A molecule of hydrogen has two hydrogen atomsbonded together and is the smallest particle of hydrogen element which hasindependent existence.
(ii) A molecule of oxygen has two atoms and is the smallest particle of oxygen elementwhich can exist independently. A molecule of water is a compound obtained fromtwo hydrogen atoms and one oxygen atom.
12
(b) An ion is a charged particle. It can be either positively charged or negativelycharged. A negatively charged ion is called an anion and a positively charged ionis called a cation. Such a group of atoms carrying a net charge is known as apolyatomic ion. For example, Na+ is a single charged atom and is known as sodium
cation. Cl– is a single charged atom and is known as chloride ion. –3NO is a group
of atoms carrying net negative charge and is known as polyatomic anion.
Symbol of sodium ion is Na+
Symbol of oxide ion is O2–
Symbol of polyatomic anion is (X)n– where X is a polyatomic ion.
Symbol of polyatomic cation is (X)n+ where X is a polyatomic ion.
Chemical Formula : Chemical formulae of a compound is a symbolic representation of itscomposition.
The chemical formulae of a molecular compound is written as follows :
1. Write the symbols of the constituent elements side by side, in such a way that the cation ison the left and the anion is on the right.
2. Write their valency numbers over the symbols and criss-cross the valency number to wirteas subscripts to the symbols.
For example : Formula of Hydrogen Chloride
H Cl
1 1
Thus, the formula of the compound will be H1 Cl1 or HCl.
Formulae of hydrogen sulphide
H S
1 2
Symbol
Valency
The formula of the compound will be H2S1 or H2S.
Formula of calcium oxide
Ca O
2+ 2–
Symbol
ChargeFormula : Ca2O2 or CaO
Writing Chemical Formulae
Assignment Based on Topic — 1 . 4
13
VERY SHORT ANSWER TYPE QUESTIONSQ. 1. Which of the following represents a correct chemical formula ? Name it.
(a) CaCl (b) BiPO4
(c) NaSO4 (d) NaS [NCERT Exemplar, H H H H ]Ans. (b) BiPO4 (Bisumth phosphate).
Here both the ions are trivatent (i.e., B + PO4)Q. 2. What is meant by the term chemical formula ? [H H H H ]Ans. The chemical formula of a compound is a symbolic representation of its composition
and actual number of atoms in one molecule of a pure substance, may be an atom or acompound.
Q. 3. How do you define the valency of an element in a molecular compound ? [H H H H ]Ans. In a molecular compound, valency of an element is defined as its combining capacity and
is equal to the number of hydrogen atoms or number of chlorine atoms or double thenumber of oxygen atoms with which one atom of the element combines.
Q. 4. How do you define the valency of an ion in an ionic compound ?[HOTS, SA-II, H H H H H ]
Ans. The valency of an ion is defined as the units of positive or negative charge present on theion.
Q. 5. Write the formula of the compound formed by the ions Al3+ and SO42–.
[CBSE 2014 C, H H H H ]Ans. Al2 (SO4)3
SHORT ANSWER TYPE QUESTIONS [TYPE—I]
Q. 1. Write down the formulae of : [H H H ](i) Sodium oxide (ii) Aluminium chloride(iii)Sodium Sulphide (iv) Magnesium hydroxide
Ans. (i) Na+1 + O2– (ii) Al3+ + Cl–
N a+ 1 O2 –
N a2 O
A l 3 + C l–
A l C l3
(iii) Na+ + S2– (iv) Mg2 + OH–
N a+ 1 S2 –
N a2 S
M g 2 + O H –
M g (O H )2
Q. 2. (a) Write chemical formulae of the following compounds :(i) Aluminium nitride (ii) Ammonium phosphate(b) Name an element which shows a variable valency. Write the formulae of its two
chlorides. [CBSE 2012 C, SA-II, H H H H H ]Ans. (a) (i) AlN (ii) (NH4)3 PO4
(b) Irron (Fe). It forms FeCl2 and FeCl3 with chlorine.
14
Q. 3. If the calcium salt of a hypothetical anion Z has the molecular formula Ca3z2. What isthe valency of z and what would be molecular formula of the aluminium salt of z ?
[HOTS, H H H H ]Ans. Valency of z = – 3
Valency of Al = + 3Formula of aluminium salt of z = Alz
Q. 4. The chemical formula for calcium oxide is CaO and not Ca2 O2. Comment.[SA-II, H H H H H ]
Ans. The chemical formula of calcium oxide is CaO because in writing the formula of acompound, valencies are divided by highest common factor.
Q. 5. Write the molecular formulae of all the compounds that can be formed by thecombination of following ions :Cu2+, Na+, Fe3+, Cl–, SO4
2–, PO43– [NCERT Exemplar]
Ans. Cation Formula of the compounds
Cu2+ CuCl2, CuSO4, Cu3 (PO4)2
Na+ NaCl, Na2SO4, Na3PO4
Fe3+ FeCl3, Fe2 (SO4)3, FePO4
SHORT ANSWER TYPE QUESTIONS [TYPE—II]Q. 1. Write the molecular formulae for the following compounds :
(a) copper (II) bromide (b) aluminium (III) nitrate(c) calcium (II) phosphate (d) Iron (III) sulphide(e) mercury (II) chloride (f) magnesium (II) acetate
[NCERT Exemplar, H H H H ]Ans (a) CuBr2 (b) Al (NO3)3
(c) Ca3 (PO4)2 (d) Fe2S3(e) HgCl2 (f) Mg (CH3 COO)2
Q. 2. Write down the names of the compounds represented by the following :(i) Al2 (SO4)3 (ii) MgCl2 (iii) K2SO4 (iv) KNO3 (v) CaCO3 (vi) (NH4)2 CO3
[HOTS, SA-II, H H H H H ]Ans. (i) Al2 (SO4)3 – Aluminium Sulphate
(ii) MgCl2 – Magnesium chloride(iii) K2SO4 – Potassium sulphate(iv) KNO3 – Potassium nitrate(v) CaCO3 – Calcium carbonate(vi) (NH4)2 CO3 – Ammonium carbonate
Q. 3. An element X has a valency :(a) Write the chemical formula of its phosphide.(b) Write the chemical formula of its chloride.(c) Is element X a metal or a non-metal ? [CBSE 2013, SA-II, H H H H H ]
Ans. (a) Chemical formula of its phosphide is X3P(b) The formula of its chloride is X Cl(c) The element X is a metal.
15
LONG ANSWER TYPE QUESTIONSQ. 1. Give the chemical formulae for the following compounds and compute the ratio by
mass of the combining elements in each one of them :(a) Ammonia (b) Carbon monoxide(c) Hydrogen chloride (d) Aluminium fluoride(e) Magnesium sulphide [H H H H ]
Ans. Compound
Name Formula No of atoms Mass ratio
(a) Ammonia NH3 1 : 3 14 : 1 × 3 = 14 : 3(b) Carbon monoxide CO 1 : 1 12 : 16 = 3 : 4(c) Hydrogen chloride HCl 1 : 1 1 : 35.5 = 2 : 71(d) Aluminium fluoride AlF3 1 : 3 27 : 3 × 19 = 9 : 19(e) Magnesium Sulphide Mg S 1 : 1 24 : 32 = 3 : 4
Q. 2. What is meant by atomicity ? How many different types of molecules are known on thebasis of atomicity ? [SA-II, H H H H ]
Ans. The number of atoms present in one molecule of the substance is called its atomicity.Depending on the number of atoms present in a molecule, the following types of moleculesare known :(i) Mono atomic molecules : The noble gases exist as single atoms and, therefore, their
molecules are monoatomic.(ii) Diatomic molecules : The molecules consisting of two atoms, for example, O2, N2,
Cl2, H2, Br2 etc.(iii) Triatomic molecules : The molecules consisting of three atoms, for example, O3 [Ozone
molecule](iv) Testatomic molecules : The molecules consisting of four atoms, for example, P4
(Phosphorus molecule)(v) Poly atomic molecules : The molecules consisting of more than four atoms, for
example, S8 (Sulphur molecule).
Molecular Mass : The molecular mass of a substance is the sum of the atomic masses of allatoms in a molecule of the substance.
For example : Molecular mass of water is 18.H2O = 2 × H × 1 × O
= 2 × 1 × 1 × 16 = 18 amu or .Atomic mass unit (amu or µ) : The mass of 1/12 part of C – 12 is equivalent to one atomic mass
unit. Previously, it was denoted by symbol amu but now, these days it is denoted by symbol .
Molecular Mass and Mole Concept
Assignment Based on Topic — 1. 5
16
Mole concept : 1 mole of a compound has a mass equal to its relative molecular mass expressedin grams.
1 mole = 6.022 × 1023 number= Relative mass in grams.
The mole (or mole) is the SI unit of amount of a substance. One mole of any species (atoms,molecules, ions or particles) is that quantity in number having mass equal to its atomic or molecularmass in grams.
The number of particles present in 1 mole of any substance is fixed, equal to 6.022 × 1023. This isa constant called the Avogadro’s number (No) or Avogadro’s constant (NA = 6.02 × 1023).
VERY SHORT ANSWER TYPE QUESTIONSQ. 1. What is meant by molar mass ? [CBSE 2015, SA-II, H H H H H ]
Ans. The mass of 1 mole of the substance (i.e., Avogadro’s number of particles) is called molarmass of the substance.
Q. 2. Define the atomic mass unit. [CBSE 2013, SA-II, H H H H H ]
Ans. One atomic mass unit is a mass unit equal to exactly one-twelfth (1/12) the mass of oneatom of carbon – 12.
Q. 3. Define mole. [HOTS, SA-II, H H H H H ]
Ans. A mole of particles (atoms, ions or molecules) is defined as that amount of the substancewhich contains the same number of particles as there are C–12 atoms in 12g i.e., 0.012 kgof C–12 istope.
Q. 4. What is the mass of 1 mole of nitrogen atoms ? [SA-II, H H H H H ]
Ans. 1 mole of nitrogen atom = 1 × gram atomic mass of nitrogen atom.= 1 × 14 g = 14 g.
Q. 5. By what name the quantity of any material that contains 6.0 × 1023 of chemical units, iscalled ? [H H H H ]
Ans. The quantity of material that contains 6.02 × 1023 chemical units is called a mole.
Q. 6. O2 gas has molar mass 32 where as SO2 gas has molar mass 64. What will be the relationbetween their molar volumes at STP ? [H H H H H ]
Ans. Gram molecular volume is the volume occupied by one mole of the substance. In case ofall gases its value is 22.4 L at STP.
Q. 7. What is the value of Avogadro’s number ? [H H H H ]
Ans. 6.022 × 1023.Q. 8. How many litres does 2 moles of H2 gas at STP contain ? [SA-II, H H H H H ]
Ans. 2 moles of H2 gas at STP contain 44.8L.Q. 9. How many mole atoms of sulphur are present in two moles of sulphuric acid, H2SO4?
[H H H ]Ans. 2 moles.
Q. 10. How many molecules are present in 4g of H2O ? [H H H H H ]
Ans.12
× 6.023 × 1023 = 3.011 × 1023 molecules.
17
SHORT ANSWER TYPE QUESTIONS [TYPE—I]Q. 1. Calculate the number of moles of magnesium present in a magnesium ribbon weighing
12g. Molar atomic mass of magnesium is 24 g mol–1. [NCERT Exemplar, H H H H ]
Ans. No. of moles of magnesium =Mass of magnesium
Molar atomic mass of magnesium
= –1
12g
24g mol = 0.5 mol
Q. 2. Calculate the number of moles in 5.75 g of sodium (atomic mass of sodium = 23)[CBSE 2014], H H H H H ]
Ans. Mass (m) = 5.75 g, Atomic mass = 23uMolar mass (M) = 23 g
Number of moles (n) = m/M= 8.57/23= 0.25 moles.
Q. 3. Define Avogadro’s constant. How is it related to mole ? [CBSE 2015, SA-II, H H H H H ]Ans. Avogadro’s number is actual number of molecules in one gram molecule of any substance.
Its value is 6.022 × 1023 particles.Mole contains the number of particles = Avogadro’s constant
Q. 4. Calculate the number of particles in 8 g of oxygen molecules. [SA-II, H H H H H ]Ans. 1 mole of O2 molecules = Gram molecular mass of O2
= 32 gAlso 1 mole of O2 molecules = 6.022 × 1023 molecules of O2
Thus, 22 g of O2 have molecules = 6.022 × 1023
0.8 g of O2 will have molecules =236.022 10
832
= 1.5055 × 1023 molecules.Q. 5. Calculate the mass of 1 molecule of water. [SA-II, H H H H H ]Ans. 1 mole of H2O = Gram molecular mass = 18 g
Also, 1 mole of H2O = 6.022 × 1023 moleculesThus, 6.022 × 1023 molecules of H2O have mass = 18 g
1 molecule of H2O will have mass = 2318
6.022 10 g = 2.989 × 10–23 g
SHORT ANSWER TYPE QUESTIONS [TYPE—II]Q. 1. A gas jar contains 1.7 g of ammonia gas. Calculate the following :
(i) Molar mass of ammonia(ii) How many moles of ammonia are present in the gas jar ?(iii)How many molecules of ammonia are present in the sample ? [CBSE 2014]
(Atomic mass N = 14, H = 1) [SA-II, H H H H H ]
18
Ans. (i) Molar mass of NH3 = 14 + 3 × 1 = 14 + 3= 17g
(ii) No. of moles =Given massMolar mass
=1.7 117 10 = 0.1 moles
(iii) Number of molecules =236.022 10 Given mass
Molar mass
=236.022 10 1.7
17
= 6.022 × 1022
Q. 2. A sample of ethane (C2H6) gas has the same mass as 1.5 × 1020 molecules of methane(CH4). How many C2H6 molecules does the sample of gas contain ? [NCERT Exemplar]
Sol. Mass of 1.5 × 1020 molecules of CH4 = 20423 –1
Molar mass of CH1.5 10
6.022 10 mol
=20
23 –1
16g/mol × 1.5 × 10
6.022 10 mol = 3.98 × 10–3g
Number of C2H6 molecules in 3.98 × 10–3g =3 23 –1
2 6
3.98 10 g 6.022 10 molmolar mass of C H
=3 23 –1
2 6
3.98 10 g 6.022 10 molmolar mass of C H
= 8 × 1019
Q. 3. Find. [SA-II, H H H H H ](a) Number of molecules of 90 g of H2O(b) Number of moles in 19 g of H2O2
(c) Formula unit mass of Al2 (CO3)3
[Atomic mass Al = 27u, C = 12u, O = 16u, H = 1u, NA = 6.022 × 1023 mol–1]Ans. (a) Molar mass of H2O = 18 g (1 mol)
18 g H2O contains 6.022 × 1023 molecules
90 g H2O contain236.022 10 90
18
= 30.110 × 1023
= 3.011 × 1024 molecules(b) Molar mass of H2O2 = 2 × 1 + 2 × 16 = 34 g 34 g of H2O2 = 1 mole of H2O2
19 g of H2O2 =1
1934 = 0.56 mol.
(c) Al2 (CO3)3 = 2 × 27 + 3 × 12 + 9 × 16= 54 + 36 + 144 = 234 g
19
LONG ANSWER TYPE QUESTIONS
Q. 1. Which has more number of atoms ? 100 g of N2 or 100g of NH3 ?[NCERT Exemplar, H H H H ]
Ans. Number of atoms in 100g of N2 = 2 × Number of molecules in 100g of N
= 2 A
2
2 mass of N Nmolar mass of N
=23 –12 100g 6.022 10 mol
28g/mol
= 43 × 1023
Number of atoms in 100g of NH3
= 4 × number of molecules in 100 g of NH3
= 3 A
3
4 Mass of NH NMolar mass of NH
= 23 –1
–1
4 100 6.022 10 mol
17 mol
g
g
= 141.7 × 1023
From the calculations, 100 g NH3 will contain more atoms than 100g of N2
Q. 2. (a) Calculate the mass of 0.5 moles of sugar (C12 H22 O11) [SA-II, H H H H H ]
[Atomic mass of C = 12, H = 1, O = 16]
(b) The actual mass of a single carbon atom is 1.992 × 10–23 grams. Compute the numberof carbon atoms in 12 grams of carbon.
(c) Chlorophyll, the green colouring matter of plants, responsible for photosynthesiscontains 2.64% of magnesium by mass. Calculate the number of magnesium atomsin 2.00 g of chlorophyll.
(d) Find the mass of one molecule of water.
(e) If 16.26 mg of a sample of an element ‘X’ contains 1.66 × 1020 atoms. What is theatomic mass of the element ‘X’ ?
Ans. (a) Molar mass of C12H22O11 = 12 × 12 + 22 × 1 + 11 × 16
= 342g
1 mole of sugar = 342g of sugar
0.5 mole of sugar = 342 × 0.5 = 171 g
(b) 1.992 × 10–23 g is the mass of 1 carbon atom.
1g will be the mass of 231
1.992 10 carbon atoms.
12g will be the mass of 2312
1.992 10 carbon atoms = 6.023 × 1023 carbon atoms.
20
(c) 100g of chlorophyll has magnesium = 2.64 g
2.0 g of chlorophyll has magnesium =2.64 2
100
= 5.28 × 10–2 g
24 g magnesium contain = 6.022 × 1023 atoms
5.28 × 10–2 g magnesium contains = 23 –26.022 10 5.28 10
24= 1.32 × 1021 atoms of Mg
(d) Molar mass of water (H2O) = 2 × 1 + 16 = 18 g 1 mol = 6.022 × 1023 molecules
= 18 g of water
1 molecule weigh = 2318
6.022 10 = 2.99 × 10–23 g
(e) 1 mole of element ‘X’ contains = 6.022 × 1023 atoms 1.66 × 1020 atoms weigh 16.26 mg = 16.26 × 10–3g
6.022 × 1023 atoms will weigh =3
2320
16.26 106.022 10
1.66 10
= 59 g
Therefore, atomic mass of element ‘X’ is 59u.
NCERT TEXTBOOK QUESTIONS
INTEXT QUESTIONS (Page No. 32–33)
Q. 1. In a reaction, 5.3g of sodium carbonate reacted with 10 g of ethanoic acid. The productswere 2.2 g of carbon dioxide, 0.9g water and 8.2 g of sodium ethanoate. Show that theseobservations are in agreement with the law of conservation of mass. [H H H H ]Sodium carbonate + ethanoic acid Sodium ethanoate + carbon dioxide + water.
Ans. Mass of reactants = Mass of sodium carbonate + Mass of ethanoic acid.= 5.3g + 6g= 11.3g
Mass of products = Mass of sodium ethanoate + Mass of carbon dioxide + Mass of water= 8.2 g + 2.2 g + 0.9g = 11.3
The mass of reactants is equal to the mass of products.Hence, it proves law of conservation of mass.
Q. 2. Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. What mass ofoxygen gas would be required to react completely with 3g of hydrogen gas. [H H H H ]
Ans. Hydrogen : OxygenRatio by mass 1 : 8
3g : ?Based on the ratio proposition.
21
Mass of oxygen required =8 3g
1
= 24 g
Q. 3. Which postulate of Dalton’s atomic theory is the result of the law of conservation ofmass ? [SA-II, H H H H H ]
Ans. Atoms are indivisible particles, which cannot be created nore be destroyed in a chemicalreaction.
Q. 4. Which postulate of Dalton’s atomic theory can explain the law of definiteproportions ? [H H H H H ]
Ans. Atoms combine in the ratio of small whole numbers to from compounds.
INTEXT QUESTIONS (Page No. 35)
Q. 1. Define the atomic mass unit. [H H H H H ]
Ans. The mass equal to1
12th of the mass of a 12
6 C atom is called one atomic mass unit.
Q. 2. Why is it not possible to see an atom with naked eyes ? [H H H H ]Ans. An atom is an extremely small particle whose radius is of the order of 10–10 m. This size is
so small that our eyes are not able to see it.
INTEXT QUESTIONS (Page No. 39)
Q. 1. Write down the formulae of :(i) Sodium oxide (ii) Aluminium chloride(iii)Sodium sulphide (iv) Magnesium hydroxide [H H H H ]
Ans. (i) Sodium oxide
N a O
+ 1 – 2
Formula = Na2O(ii) Aluminium Chloride
A l C l
+3 –1
Formula = AlCl3
(iii) Sodium Sulphide
N a S
+ 1 – 2
Formula = Na2S(iv) Magnesium hydroxide
22
M g (O H )
+2 –1Formula = Mg (OH)2
Q. 2. Write down the names of compounds represented by the following formulae :(i) Al2 (SO4)3 (ii) CaCl2
(iii)K2SO4 (iv) KNO3 [H H H H ](v) CaCO3
Ans. (i) Aluminium sulphate (ii) Calcium chloride(iii) Potassium sulphate (iv) Potassium nitrate(v) Calcium carbonate
Q. 3. What is meant by the term chemial formula ? [H H H H ]Ans. It is a symbolic representation of its composition.Q. 4. How many atoms are present in a :
(i) H2S molecule and (ii) PO43– ion ? [H H H H ]
Ans. (i) 3 atoms are present in H2S.(ii) 5 atoms are present in PO4
3– ion.
INTEXT QUESTIONS (Page No. 40)
Q. 1. Calculate the molecular masses of H2, O2, Cl2, CO2, CH4, C2H6, C2H4, NH3, CH3OH.[At mass of H = 1, O = 16, Cl = 35.5, C = 12, N = 14] [H H H H H ]
Ans. (i) Molecular mass of H2 = 2 × atomic mass of H= 2 × 1u = 2u
(ii) Molecular mass of O2 = 2 × atomic mass of O= 2 × 16u = 32u
(iii) Molecular mass of Cl2 = 2 × atomic mass of Cl= 2 × 35.5 u = 71u
(iv) Molecular mass of CO2 = 1 × atomic mass of C + 2 × atomic mass of O= 1 × 12u + 2 × 16u = 44u
(v) Molecular mass of CH4 = 1 × atomic mass of C + 4 × atomic mass of H= 1 × 12u + 4 × 1u = 16u
(vi) Molecular mass of C2H6 = 2 × atomic mass of C + 6 × atomic mass of H= 2 × 12u + 6 × 1u = 30u
(vii) Molecular mass of C2H4 = 2 × atomic mass of C + 4 × atomic mass of H= 2 × 12u + 4 × 1u = 28u
(viii) Molecular mass of NH3 = 1 × atomic mass of N + 3 × atomic mass of H= 1 × 14u + 3 × 1u = 17u
(ix) Molecular mass of CH3OH = 1 × atomic mass of C + 4 × atomic mass of H + 3× atomic of O
= 1 × 12u + 4 × 1u + 1 × 16u = 32u
23
Q. 2. Calculate the formula unit masses of ZnO, Na2O, K2CO3 given atomic masses of Zn =65u, Na = 23u, K = 39u, C = 12u, and O = 16u.
Ans. (i) Formula unit mass of ZnO = 1 × atomic mass of Zn + 1 × atomic mass of O= 1 × 65u + 1 × 16u = 81u
(ii) Formula unit mass of Na2O = 2 × atomic mass of Na + 1 × atomic mass of O= 2 × 23u + 1 × 16u = 62u
(iii) Formula unit mass of K2CO3= 2 atomic mass of k + 1 × atomic mass of C + 3× atomic mass of O
= 2 × 39u + 1 × 12u + 3 × 16u= 78u + 12u + 48u = 138u
INTEXT QUESTIONS (Page No. 42)
Q. 1. If one of carbon atoms weighs 12 grams, what is the mass (in gram) of 1 atom ofcarbon ?
Ans. 1 mole i.e., 6.02 × 1023 atoms of carbon weighs = 12g
1 atom of carbon weighs = 2312
6.02 10 = 1.99 × 10–23g
Q. 2. Which has more number of atoms, 100 grams of sodium or 100 grams of iron (givenatomic mass of Na = 23u, Fe = 56u) ?
Ans. 23 g of sodium has atoms = NA (Avogadro’s number)
100 g of sodium has atoms =10023 NA = 4.3 NA atoms
Also, 56 g of iron has atoms = NA (Avogadro’s number)
100g of iron has atoms =10056 NA = 1.78 NA atoms
Hence, 100g of sodium has more atoms than 100g of iron.
NCERT TEXTBOOK EXERCISE
Q. 1. A 0.24 g sample of compound of oxygen and boron was found by analysis to contain0.096 of boron and 0.144 g of oxygen. Calculate the percentage composition of thecompound by weight. [SA-II, H H H H H ]
Sol. Mass of the compound = 0.24 gMass of boron = 0.096 g
Mass of oxygen = 0.144 g
Percentage of boron =Mass of boron
100Mass of compound
=0.096g
1000.240g
= 40%
24
Percentage of oxygen =Mass of oxygen
100Mass of compound
=0.144
1000.240
gg = 60%
Q. 2. When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced.What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g ofoxygen. Which law of chemical combination will govern your answer ? [H H H H ]
Ans. Carbon + Oxygen Carbon dioxide
3.0 g 8.00 g 11.00 g
3.0 g carbon + 8 g oxygen = 11 g CO2
If 50 g of oxygen is there, only 8 g of O2 will react with 3.0 g of carbon.
(50 – 8) = 42 g of oxygen will remain unreacted
The above data tells that the reaction between carbon and oxygen is governed by law ofconstant proportions.
Q. 3. What are poly atomic ions ? Give examples ? [SA-II, H H H H ]
Ans. The ions which contain more than one atom and behave as a single unit are calledpolyatomic ions.
Example : Ammonium ions (NH4+), carbonate ion (CO3
2–).
Q. 4. Write the chemical formulae of the following : [SA-II, H H H H H ]
(a) Magnesium chloride (b) Calcium oxide
(c) Copper nitrate (d) Aluminium chloride
(e) Calcium carbonate
Ans. (a) Magnesium chloride
M g C l
2+ 1–
Sym bol
C harge
= M gC l2
(b) Calcium oxide
C a O
2+ 2–
Sym bol
C harge
= C aO
(c) Copper nitrate
C u (N O )3
2+ 1–
Sym bol
C harge
= Cu (N O )3 2
25
(d) Aluminium chloride
A l C l
3+ 1–
Sym bol
C harge
= A lC l3
(e) Calcium carbonate
C a (CO )3
2+ 2–
Sym bol
C harge
Dividing by the common factor (2) and applying crisis – cross. Formula = CaCO3
Q. 5. Give the names of the elements present in the following compounds : [H H H H ](a) Quick lime (b) Hydrogen bromide(c) Baking powder (d) Potassium sulphate
Ans. (a) Quick lime is calcium oxide (CaO)Elements present : Calcium (Ca) and oxygen (O)
(b) Hydrogen bromide is HBr.Elements present : Hydrogen (H) and bromine (Br)
(c) Baking powder is sodium hydrogen carbonate (NaH CO3).Elements present : Sodium (Na), hydrogen (H), Carbon (C), and Oxygen (O)
(d) Potassium sulphate (K2SO4)Element present : Potassium (k), Sulphur (S) and Oxygen (O)
(e) Calcium carbonate is (CaCO3)Element present : Calcium (Ca), Carbon (C) and Oxygen (O)
Q. 6. Calculate the molar mass of the following substances : [SA-II, H H H H ](a) Ethyne(b) Sulphur molecule, S8
(c) Phosphorus molecule, P4 (Atomic mass of phosphorus = 31)(d) Hydrochloric acid, HCl(e) Nitric acid, HNO3
Ans. (a) Molar mass of ethyne, (C2H2) = 2 × atomic mass of C + 2 × atomic mass of H= 2 × 12u + 2 × 1u = 26u
(b) Molar mass of sulphur, (S8) = 8 × atomic mass of S= 8 × 32u = 256u
(c) Molar mass of hydrochloric acid,(HCl) = 1 × atomic mass of H + 1 × atomic mass of Cl
= 1 × 1u + 35.5 u = 36.5u(d) Molar mass of nitric acid,
(HNO3) = 1 × atomic mass of H + 1 × atomic mass of N+ 3 × atomic mass of O
= 1 × 1u + 1 × 14u + 3 × 16u= 63u.
26
Q. 7. What is the mass of : [H H H H H ](a) 1 mole of nitrogen atom ?(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27) ?(c) 10 moles of sodium sulphite (Na2 SO3) ?
[Atomic mass N = 14u, Al = 27u, Na = 23u, S = 32u, O = 16u] ?Ans. (a) 1 mole of nitrogen atoms = 1 × gram atomic mass of nitrogen atom
= 1 × 14g = 14g(b) 4 moles of aluminium atoms = 4 × gram atomic mass of aluminium atoms
= 4 × 27g = 108g(c) 10 moles of sodium sulphite (Na2 CO3)
= 10 (2 × gram atomic mass of Na + 1× gram atomic mass of sulphur + 3× (gram atomic mass of oxygen)
= 10 (2 × 23g + 1 × 32g + 3 × 16g)= 10 (46g + 32g + 48g)= 10 × 126g = 1260g.
Q. 8. Convert into mole : [SA-II, H H H H H ](a) 12 g of oxygen gas (b) 20 g of water(c) 22 g of carbond ioxide
Ans. (a) 1 mole of oxygen (O2) = 2 × 16 g = 32 g 32 g oxygen = 1 mole
12 g of oxygen =1232 mole = 0.375 mole
(b) 1 mole of water (H2O) = 2 × 1g + 1 × 16 g = 18 g 18 g of water = 1 mole
20 g of water =2018
mole = 1.11 mole
(c) 1 mole of carbon dioxide (CO2) = 1 × 12 g + 2 × 16 g = 44 g 44 g of carbon dioxide = 1 mole
22 g of carbon dioxide =2244
mole = 0.5 mole
Q. 9. What is the mass of : [SA-II, H H H H H ](a) 0.2 mole of oxygen atoms ?(b) 0.5 mole of water molecules ?
Ans. (a) Mass of 0.2 mole atoms of oxygen = 0.2 mol × molar mass of oxygen atom= 0.2 mol × 16g/mol= 3.2 g
(b) Mass of 0.5 mole molecules of water= 0.5 mol × molar mass of water= 0.5 mol × 18 g 1 mol = 9 g.
27
Q. 10. Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.(At mass Al = 27u, O = 16u) [H H H H ]
Ans. Molecular formula of aluminium oxide = Al2O3
Molecular mass of Al2O3 = 2 × 27 + 3 × 16= 54 + 48 = 102g mol–1
Number of moles =Given mass
Formula unit mass
= –1
0.051g
102g mol = 0.0005 mole
= 5 × 10–4 moleOne mole of Al2O3 contains 6.022 × 1023 formula unit of Al2O3
5 × 10–4 mole of Al2O3 will contain formula units of Al2O3
= (5 × 10–4) × (6.022 × 1023)= 30.11 × 1019
= 3.011 × 1020
Now, one formula unit of Al2O3 contains 2 aluminium ions. 3.011 × 1020 formula units will contain aluminium ions
= 2 × 3.011 × 1020
= 6.022 × 1020
NCERT EXEMPLAR PROBLEMS
MULTIPLE CHOICE QUESTIONS1. Which of the following correctly represents 360 g of water ? [H H H H ]
(i) 2 moles of H2O(ii) 20 moles of water(iii) 6.022 × 1023 molecules of water(iv) 1.2044 × 1025 molecules of water(a) (i) (b) (i) and (iv)(c) (ii) and (iii) (d) (ii) and (iv)
Ans. (d) (ii) and (iv)2. Which of the following statement is not true about an atom ? [SA-II, H H H H H ]
(a) Atoms are not able to exist independently(b) Atoms are the basic units from which molecules and ions are formed(c) Atoms are always neutral in nature(d) Atoms aggregate in large numbers to form the matter that we can see, feel or touch.
Ans. (a) Atoms are not able to exist independently.
28
3. The chemical symbol for nitrogen gas is : [H H H H ](a) Ni (b) N2
(c) N+ (d) NAns. (b) N2
4. The chemical symbol for sodium is : [H H H ](a) So (b) Sd(c) NA (d) Na
Ans. (d) Na5. Which of the following would weigh the highest ? [SA-II, H H H H H ]
(a) 0.2 mole of sucrose (C12 H22 O11) (b) 2 moles of Co2
(c) 2 moles of CaCO3 (d) 10 moles of H2OAns. (c) 2 moles of CaCO3
6. Which of the following has maximum number of atoms ? [H H H H ](a) 18 g of H2O (b) 18 g of O2
(c) 18 g of CO2 (d) 18 g of CH4
Ans. (d) 18 g of CH4
7. Which of the following contains maximum number of molecules ? [SA-II, H H H H H ](a) 1g CO2 (b) 1g N2
(c) 1g H2 (c) 1g CH4
Ans. (a) 1g CO2
8. Mass of one atom of oxygen is :
(a) 2316
g6.023 10
(b) 2332
g6.023 10
(c) 231
g6.023 10
(d) 8u
Ans. (a) 2316
g6.023 10
9. 3.42 g of sucrose are dissolved in 18g of water in a beaker. The number of oxygenatoms in the solution are : [H H H H ](a) 6.68 × 1023 (b) 6.09 × 1022
(c) 6.022 × 1023 (d) 6.022 × 1021
Ans. (a) 6.68 × 1023
10. A change in the physical state can be brought about : [H H H H H ](a) only when energy is given to the system(b) only when energy is taken out from the system.(c) when energy is either given to, or taken out from the system.(d) without any energy change.
Ans. (c) when energy is either given to, or taken out from the system.
29
SHORT ANSWER TYPE QUESTION11. Which of the following represents correct chemical formula ? Name it. [H H H ]
(a) CaCl (b) BiPO4
(c) NaSO4 (d) NaSAns. Correct answer is (b) BiPO4
Here both the ions are trivalent (i.e., Bi3+ PO43–)
12. Write the molecular formulae for the following compounds : [SA-II, H H H H ](a) copper (ii) bromide (b) aluminium (iii) nitrate(c) calcium (ii) phosphate (d) iron (iii) sulphide(e) mercury (ii) chloride (f) magnesium (ii) acetate
Ans. (a) CuBr2 (b) Al (NO3)(c) Ca3 (PO4)2 (d) Fe2S3
(e) HgCl2 (f) (CH3 COO)2 Mg13. Write the molecular formulae of all the compounds that can be formed by the
combination of following ions : [H H H H ]Cu2+, Na+, Fe3+ Cl–, SO4
2–, PO43–
Ans. Cation Formulae of the compoundsCu+ CuCl2, CuSO4, Cu3 (PO4)2
Na+ NaCl, Na2SO4, Na3PO4
Fe3+ FeCl3, Fe2(SO4)3, FePO4
14. Write the cations and anions present (if any) in the following compounds :(a) CH3COONa (b) NaCl(c) H2 (d) NH4NO2 [SA-II, H H H H H ]
Ans. Compound Cation Anion
(a) CH3COONa Na+ CH3COO–
(b) NaCl Na+ Cl–
(c) H2 –– ––(covalent compound, no ions)
(d) NH4NO3 NH4+ NO3
–
15. Give the formulae of the compounds formed from the following sets of elements :(a) Calcium and flucorine (b) Hydrogen and sulphur(c) Nitrogen and hydrogen (d) Calcium and chlorine [H H H H ](e) Sodium and oxygen (f) Carbon and oxygen
Ans. (a) Calcium fluoride, CaF2 (b) Hydrogen sulphide, H2S(c) Ammonia, NH3 (d) Carbon tetra chloride, CCl4
(e) Carbon dioxide, CO2 (f) Carbon monoxide, CO16. Which of the following symbols of elements are incorrect ? Give their correct symbols.
(a) Cobalt, CO (b) Carbon, C(c) Aluminium AL (d) Helium, He(e) Sodium, So [H H H ]
30
Ans. Correct/Incorrect Correct symbols(a) Incorrect Co(b) Incorrect C(c) Aluminium Al(d) Helium He(e) Incorrect Na
17. Give the chemical formulae for the following compounds and compute the ratio bymass of the combining elements in each one of them. [SA-II, H H H H H ](a) Ammonia (b) Carbon monoxide(c) Hydrogen chloride (d) Aluminium fluoride(e) Magnesium sulphide
(Atomic mass N = 14 u, H = 1 u, C = 12, O = 16 u, Cl = 35.5 u, F = 19 u, Mg = 24 u,and S = 32 u)
Ans. Name of the compound Formula No. of atoms Mass ratio
(a) Ammonia NH3 1 : 3 14 : 1 × 3 = 14 : 3(b) Carbon monoxide CO 1 : 1 12 : 16 = 3 : 4(c) Hydrogen chloride HCl 1 : 1 1 : 35.5 = 2 : 71(d) Aluminium florid Al F3 1 : 3 27 : 3 × 19 = 9 : 19(e) Magnesium sulphide MgS 1 : 1 24 : 32 = 3 : 4
18. State the number of atoms present in each of the following chemical species :
(a) 2–3CO (b) 3–
4PO
(c) P2O5 (d) CO [SA-II, H H H H H ]Ans. (a) 1 + 3 = 4 (b) 1 + 4 = 5
(c) 2 + 5 = 7 (d) 1 + 1 = 219. What is the fraction of the mass of water due to neutrons ? [SA-II, H H H H ]
Ans. Molecular mass of water (H2O) = 2u + 16 u = 18 uMass of neutrons in water molecule = 8 u
Therefore,fraction of mass of water due to neutrons=8 4
18 9uu = 0.44
20. Does the solubility of a substance change with temperature ? Explain with the helpof an example. [H H H H ]
Ans. Yes. In general solubility of a substance change with temperature. In most cases, solubilityof a solid substance increases with a rise in temperature. Solubility of NH4Cl in water at10 °C is 24 g/100 g water and at 60 °C, it is 55 g/100 g water.
21. Classify each of the following on the basis of their atomicity. [SA-II, H H H ](a) F2 (b) NO2 (c) N2O(d) C2H6 (e) P4 (f) H3O2
(g) P4O10 (h) O3 (i) HCl(j) CH4 (k) He (l) Ag
Ans. (a) Diatomic (b) Triatomic (c) Triatomic(d) Octatomic (e) Tetratomic (f) Tetratomic
31
(g) Tetradecatomic (h) Triatomic (i) Diatomic(j) Pentatomic (k) Monoatomic (l) Monoatomic
22. You are provided with a fine white coloured powder which is either sugar or salt.How would identify it without tasting ? [SA-II, H H H H ]
Ans. Bring a small amount of the white powder into the flame. If a golden yellow flame isobserved, it is NaCl (salt) and it is burns with a smell of burnt sugar, it is sugar.
23. Calculate the number of moles of magnesium present in a magnesium ribbon weighing12 g. Molar atomic mass of magnesium is 24 g mol–1. [H H H H ]
Ans. 1 mole of Mg = 24 g of Mgor 24 g of Mg = 1 mole of Mg
12 g of Mg =124
× 12 mole = 0.5 mole.
LONG ANSWER TYPE QUESTIONS24. Verify by calculating that [H H H H H ]
(a) 5 moles of CO2 and 5 moles of H2O do not have the same mass.(b) 240 g of calcium and 240 g magnesium elements have a mole ratio of 3 : 5.
[Atomic mass; C = 12, O = 16, H = 1, Ca = 40, Mg = 24]Ans. (a) Molar mass of CO2 = 12 + 2 × 16 = 44 g mol–1
or 1 mole of CO2 = 44 g 5 moles of CO2 = 5 × 44 g = 220 g
Molar mass of H2O = 2 × 1 + 16 = 18 g mol–1
or 1 mole of H2O = 18 g 5 moles of H2O = 5 × 18 g = 90 gThus, 5 moles of CO2 and 5 moles of H2O do not have the same mass.(b) Molar mass of Ca = 40 g mol–1
or 40 g of Ca = 1 mole
240 g of Ca =140
× 240 = 6 moles
Molar mass of Mg = 24 g mol–1
or 240 of Mg = 1 mole
240 g of Mg = 1
24024
= 10 moles
Mole ratio of Ca : Mg = 6 : 10 = 3 : 525. Find the ratio by mass of the combining elements in the following compounds :
(a) CaCO3 (b) MgCl2 (c) H2SO4
(d) C2H5OH (e) NH3 (f) Ca(OH)2 [H H H H ]Ans. (a) CaCO3, Ca : C : O = 40 : 12 : 3 × 16 = 40 : 12 : 48 = 10 : 3 : 12
(b) MgCl2, Mg : Cl = 24 : 2 × 35.5 = 24 : 71(c) H2SO4, H : S : O = 2 × 1 : 32 : 4 × 16 = 2 : 32 : 64 = 1 : 16 : 32(d) C2H5OH or C2H6O, C : H : O = 2 × 12 : 6 × 1 : 16 = 24 : 6 : 16 = 12 : 3 : 8(e) NH3, N : H = 14 : 3 × 1 = 14 : 3(f) Ca(OH)2, Ca : O : H = 40 : 2 × 16 : 2 × 1 = 4 : 32 : 2 = 20 : 16 : 1
32
26. Calcium chloride when dissolved in water dissociates into the ions according to thefollowing equation :
CaCl2 (aq) Ca2+ (aq) + 2Cl– (aq)Calculate the number of ions obtained from CaCl2 when 222 g of it is dissolved inwater. [Atomic mass Ca = 40 u, Cl = 35.5 u] [SA-II, H H H H H ]
Ans. Molar mass of CaCl2 = 40 + 2 × 35.5 = 40 + 71 = 111 g mol–1
or 111 g of CaCl2 = 1 mole of CaCl2
222 g of CaCl2 = 2 moles of CaCl2
= 2 × 6.022 × 1023 formula units of CaCl2
= 12.044 × 1023 formula units of CaCl2
But 1 formula unit of CaCl2 contains 1 Ca2+ ion and 2 Cl– ions 12.044 × 1023 formula units of CaCl2 will contain 12.044 × 1023 Ca2+ ions and 2 ×12.044 × 1023 or 24.088 × 1023 Cl– ions.
The ions present = (12.044 × 24.088) × 1023
= 36.132 × 1023
27. The difference in the mass of 100 moles each of sodium atoms and sodium ions is5.48002 g. Compute the mass of an electron.[H H H H ]
Ans. Na Na+ + e–
One sodium atom gives one electron.
Therefore, mass of an electron = 23 –1
5.48002 g
100mol × 6.022 × 10 mol= 9.1 × 10–28 g = 9.1 × 10–31 kg
28. Cinnabar (HgS) is a prominent ore of mercury. How many grams of mercury are presentin 225 g of pure HgS ? Molar mass of H g and S are 200.6 g mol–1 and 32 g mol–1
respectively. [H H H H ][Atomic mass = Hg = 200.6 u, S = 32 u]
Sol. Molar mass of HgS = (200.6 + 32) g mol–1
= 232.6 g mol–1
Thus, 232.6 g of HgS contain Hg = 200.6 g
225 g of HgS will contain Hg =200.6
225 g232.6
= 194.045 g
29. The mass of one steel screw is 4.11 g. Find the mass of one mole of these steel screws.Compare this value with the mass of the Earth (5.98 × 1024 kg). Which one of the two isheavier and by how many times ? [SA-II, H H H H ]
Ans. Mass of one mole of steel screws = 4.11 g × 6.022 × 1023 = 2.475 × 1024 g= 2.475 × 1021 kg.
Mass of the earth = 5.98 × 1024 kgComparison of the masses shows that the Earths is heavier than one mole of steel screws.
Ratio of masses =24
21
5.98 10 kg
2.475 × 10 kg
= 2416
Thus, the earth is 2416 times heavier than one mole of screws.
33
30. A sample of vitamin C is known to contain 2.58 × 1024 oxygen atoms. How manymoles of oxygen atoms are present in the sample ?[SA-II, H H H H ]
Ans. Number of oxygen atoms = 2.58 × 1024
Mole of oxygen atoms present =24
232.58 10
6.022 10
= 4.28
31. Raunak took 5 moles of carbon atoms in a container and Krish also took 5 moles ofsodium atoms in another container of the same weight : (a) whose container isheavier ? (b) whose container has more number of atoms ?
[Atomic mass : C = 12u, Na = 23 u] [H H H H H ]Ans. (a) 1 mole of C – atoms = 12 g
5 moles of C – atoms = 5 × 12 g = 60 g1 mole of Na-atoms = 23 g
5 mole of Na-atoms = 5 × 23 g = 115 gThus, Krish’s container is heavier.(b) 1 mole of C – atoms = 6.022 × 1023 atoms
1 mole of Na-atoms = 6.022 × 1023 atoms 5 moles of each will contain the same number of atoms.
32. Fill in the missing data in the Table : [SA-II, H H H H H ]
Species H2O CO2 Na atom MgCl2
PropertyNumber of moles 2 — — 0.5Number of particles — 3.011 × 1023 — —Mass 36 g — 115 g —
Ans. Species H2O CO2 Na atom MgCl2
PropertyNumber of moles 2 0.5 mol 5 mol 0.5Number of particles 12.044 × 1023 3.011 × 1023 30.11 × 1023 3.011 × 1023
Mass 36 g 22 g 115 g 47.5 g
33. The visible universe is estimated to contain 1022 stars. How many moles of stars arepresent in the visible universe ? [H H H ]
Ans. Number of moles of stars in the visible universe
=22
23 –110
6.022 10 mol = 0.017 mol.
34. What is the SI prefix for each of the following multiples and sub multiples of a unit?(a) 103 (b) 10–1 (c) 10–2
(d) 10–6 (e) 10–9 (f) 10–12 [SA-II, H H H H ]Ans. (a) Kilo (b) Deci (c) Centi
(d) Micro (e) nano (f) pico
34
35. Express each of the following in kilograms : [H H H H ](a) 5.84 × 10–3 mg (b) 58.34 g (c) 0.584 g(d) 5.873 × 10–21 g
Ans. (a) 5.84 × 10–3 mg = 5.84 × 10–3 × 10–3 g= 5.84 × 10–3 × 10–3 × 10–3 kg= 5.84 × 10–9 kg
(b) 58.34 g =58.34g
1000 g/kg = 5.834 × 10–2 kg
(c) 0.584 g =0.584g
1000 g/kg = 5.84 × 10–4 kg
(d) 5.873 × 10–21 g =215.873 10 g
1000 g/kg
= 5.8743 × 10–24 kg
36. Compute the difference in masses of 103 moles each of magnesium atoms andmagnesium ions. (Mass of an electron = 9.1 × 10–31 kg) [SA-II, H H H H ]
Ans. Difference of mass per atom and ion= Mass of 2 electrons= 2 × 9.1 × 10–31 kg
Therefore,Difference in the masses of 103 moles of Mg atoms
and ions = 103 mol × 2 × 9.1 × 10–31 kg= 103 × 6.022 × 1023 × 2 × 9.1 × 10–31 kg= 1.1 × 10–3 kg
37. Which has more number of atoms ? 100 g of N2 or 100 g of NH3 ? [SA-II, H H H H ]Ans. Number of atoms in 100 g of N2= 2 × number of molecules of N2
=2 A
2
2 Mass of N NMolar mass of N
=23 –12 100g × 6.022 × 10 mol
28g/mol
= 43 × 1023 mol.Number of atoms in 100 g of NH3 = 4 × Number of molecules in 100 g of NH3
= 3 A
3
4 Mass of NH NMolar mass of NH
=23 –1
–1
4 100g × 6.022 × 10 mol
17 g mol
= 141.7 × 1023
Therefore, 100 g of NH3 will contain more atoms than 100 g of N2
38. Compute the number of ions present in 5.85 g of sodium chloride. [H H H H ][Atomic mass : Na = 23 u, Cl = 35.5 u]
35
Ans. 1 mole of NaCl = 23 × 35.5 g = 58.5 g of NaCl= 6.022 × 1023 formula units of NaCl
Thus, 58.5 of NaCl have 6.022 × 1023 formula units of NaCl 5.85 g of NaCl will have NaCl formula units
=236.022 10
58.5
= 6.022 × 1022
Each formula unit of NaCl contains 1 Na+ ion and 1 Cl– ion. 6.022 × 1022 formula units of NaCl will contain 6.022 × 1022 Na+ ions and 6.022 × 1022
Cl– ions.Total ions = 2 × 6.022 × 1022
= 12.04 u × 1022
39. A gold sample contains 90% of gold and the rest copper. How many atoms of gold arepresent in one gram of this sample of gold ? [Atomic mass of Au = 197u] [SA-II, H H H H H ]
Ans. Gold sample contains 90% of gold.Therefore, 100g of the sample contains 90 g of gold
1 g of the sample will contain gold =90
100g = 0.9 g
Atomic mass of gold = 197 u 1 mole of gold atoms = 197 g
= 6.022 × 1023 atoms 197 g of gold has atoms = 6.022 × 1023
0.9 g of gold will have atoms =236.022 10
0.9197
= 0.0275 × 1023
= 2.75 × 1021 atoms.40. What are ionic and molecular compounds ? Give examples. [H H H H H ]
Ans. Ionic compounds : The compounds in which the constituent particles are ions are calledionic compounds. Example : Sodium chloride, calcium oxide etc.Molecular compounds : The compounds in which the constituent particle are discretemolecules are called molecular compounds. Example : Ammonia, carbon dioxide etc.
41. Compute the difference in masses of one mole each of aluminium atoms and one moleof its ions. (Mass of an electron is 9.1 × 10–28 g). Which one is heavier ? [H H H H H ]
Ans. Al Al3+ + 3e–
1 mol 3 mol
Differences in masses = Mass of 3 moles of electrons= 3 × 6.022 × 1023 × Mass of an electrons.= 3 × 6.022 × 1023 × 9.1 × 10–28 g= 1.64 × 10–3 g
42. A silver ornament of mass ‘m’ gram is polished with gold equivalent to 1% of the massof silver. Compute the ratio of the number of atoms of gold and silver in the ornament.
[H H H H H ]Ans. Mass of silver = Mg
36
Mass of gold =1
mg 0.01mg100
Number of atoms of silver =Am N
100 g/mol
Number of atoms of gold =A0.01m × N
197
Ratio between the number of atoms of silver and gold
=
A
A
m N 197108g/mol 0.01 m × N
= 197
1824 : 11.08
43. A sample of ethane (C2H6) gas has the same mass as 1.5 × 1020 molecules of methane(CH4). How many C2H6 molecules does the sample of gas contain ?
[Atomic mass : C = 12 u, H = 1 u] [H H H H ]
Ans. Molar mass of CH4 = 12 + 4 = 16 g mol–1
1 mole of CH4 = 16 g = 6.022 × 1023 molecules6.022 × 1023 molecules of CH4 have mass = 16 g
1.5 × 1020 molecules of CH4 will have mass =
20 3
2316
1.5 10 4 106.022 10
g
Molar mass of C2H6 = 2 × 12 × 6 = 30 g mol–1
1 mole of C2H6 = 30 g = 6.022 × 1023 molecules30 g of C2H6 have molecules = 6.022 × 1023 molecules
4 × 10–3 g of C2H6 will have molecules =
2336.022 10
4 1030
= 0.803 × 1023 = 8.03 × 1019
44. Fill in the blanks : [SA-II, H H H H H ]
(a) In a chemical reaction, the sum of the masses of the reactants and products remainunchanged. This is called ________
(b) A group of atoms carrying a fixed charge on them is called ___________
(c) The formula unit mass of Ca3 (PO4)2 is ___________
(d) Formula of sodium carbonate is ________ and that of ammonium sulphate is _____
[Atomic mass : Ca = 40 u, P = 31, O = 16]
Ans. (a) Law of conservation of mass(b) A polyatomic ion(c) 310 u [Ca3 (PO4)2 = 3 × 40 + 2 (31 + 4 × 16) = 120 + 190 = 310 u](d) Na2 CO3, (NH4)3 PO4.
37
FORMATIVE ASSESSMENT
A. Multiple Choice Questions (M.C.Q)(i) Which of the following has smallest mass ?
(a) 4g of He (b) 6.022 × 1023 atoms of He(c) 1 atoms of He (d) 1 mole atoms of He
Ans. (c) 1 atom of He(ii) The formula for quick lime is :
(a) CaCl2 (b) CaCO3
(c) Ca (OH)2 (d) CaOAns. (d) CaO(iii) The symbol of cadmium is :
(a) Ca (b) Cu(c) Cm (d) Cd
Ans. (d) Cd.(iv) The formula of ethanol is C2H5OH. Its molecular mass is :
(a) 46 u (b) 34 u
(c) 34 g (d) 46 gAns. (a) 46 u.
(v) The number of moles in 6.4 g of SO2 is :
(a) 1 (b) 0.1(c) 10 (d) 6.4
Ans. (b) 0.1(vi) Which of the following is a polyatomic ion ?
(a) Zn2+ (b) PO43–
(c) Mg2+ (d) Cr3+
Ans. (b) PO43–
(vii) A sample contains 22 g of carbon dioxide. This is equal to :
(a) one molar volume of carbondioxide (b) one mole of carbondioxide(c) half mole of carbondioxide (d) two moles of carbondioxide
Ans. (c) half mole of carbondioxide.(viii) The correct formula of aluminium sulphate is :
(a) AlSO4 (b) Al2 SO4
(c) Al3 (SO4)2 (d) Al2 (SO4)3
Ans. (d) Al2 (SO4)3
(ix) Which of the following has the lowest number of moles ?
(a) 5 g of H2O (b) 5 g of Cl2
(c) 5 g of CO2 (d) 5 g of N2
Ans. (b) 5 g of Cl2
38
(x) One mole of carbondioxide means :
(a) 44 g of carbondioxide (b) 6.022 × 1023 molecules of CO2
(c) 22.4 L at STP (d) all of the aboveAns. (d) all of the above.
B. (i) The formula of the compound containing Pb2+ and PO43– ions is ____.
(ii) The number of atoms, ions or molecules in one mole of a substance is _____.(iii) The atomiccity of ozone is _____.(iv) The number of moles in 1.8 g of H2O are ________.(v) Avogadro number is _______.(vi) 1 u stands for ______ of carbon atom.(vii) A ________ is smallest particle of an element or compound, which is capable of
independent existence.(viii) The chemical symbol used for ______ is Li(ix) Volume of 17 g of NH3 at STP is ______.(x) The formula of washing soda is ________.
Ans. (i) Pb3 (PO4)2 (ii) 6.022 × 1023 (iii) Three (iv) 0.1 (v) 6.022 × 1023 (vi)1
12th (vii) molecule
(viii) Lithum (ix) 22.4L (x) Na2CO3
C. State True / False :
(i) One mole of every substance has same mass.(ii) Mass of 1 mole of CO2 is 44 g.(iii) An atom is the smallest particle of an element or compound which is capable of
independent existence.(iv) The molecular formula for nitric acid is HNO.(v) Mass of 6.022 atoms of an element is called atomic mass.(vi) One mole of CO2 and SO2 contains same number of oxygen atoms.(vii) The elements having one valency are called monovalent.(viii) A mole always represents 6.023 × 1022 atoms or molecules or ions of a substance.(ix) Mass of two mole of atoms or molecules is called molar mass.(x) The formula of cupric oxide and cuprons oxide are CuO and Cu2O respectively.
Ans. (i) False, (ii) True, (iii) False, (iv) False, (v) False, (vi) True, (vii) True, (viii) False, (ix) False,(x) True.
D. Matching Type : [H H H H H ]
Column A Column B
1. Sulphuric acid (a) Mg (OH)2
2. Acetic acid (b) Al (OH)3
3. Magnesium hydroxide (c) CH3COOH4. Aluminium hydroxide (d) Ca (OH)2
5. Calcium hydroxide (e) H2SO4
Ans. 1 – (e), 2 – (c), 3 – (a), 4 – (b), 5 – (d).
39
1. Define the atomic mass unit. 12. What is meant by the term chemical formula ? 13. Why is it not possible to see an atom with naked eyes ? 14. How many atoms are present in a :
(i) H2S molecule and (ii) PO43– ion. 1
5. Write down the formulae of :(i) Aluminium chloride (ii) Sodium sulphide. 1
6. If one mole of carbon atoms weighs 12 gram what is the mass (in grams) of 1 atom ofcarbon ? 2
7. Write the names of the elements present in the following compounds :(a) Baking powder (b) Quick lime(c) Hydrogen bromide (d) Potassium sulphate 2
8. Which postulate of Dalton theory :(i) Is the result of the law of conservation of mass(ii) Can explain the law of definite proportions ? 2
9. Calculate the molecular mass of : (i) CO2 (ii) CH4 [Atomic mass : C = 12 u, O = 16 u,H = 1 u] 2
10. Give two differences between cation and anion. 211. What is the mass of :
(a) 1 mole of nitrogen atoms ?(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)(c) 10 moles of sodium sulphite (Na2 SO3) ? 3
12. Convert into mole :(a) 12 g of oxygen gas (b) 20 g of water(c) 22 g of carbon dioxide. 3
13. Calculate the molar mass of the following substances :(a) Ethyne, C2H2 (b) Sulphur molecule, S8
(c) Phosphorus mole, P4
[Atomic mass : C = 12 u, H = 1 u, S = 32 u, P = 31 u] 314. Calculate the mass of the following :
(i) 0.5 mole of N2 gas (ii) 0.5 mole of N atoms(iii) 3.011 × 1023 N atoms. 3
15. In a reaction, 4.2 g of sodium carbonate were reacted with 10 g of ethanoic acid solution.The products were 2.2 g of carbon dioxide and 12 g of sodium ethanoate solution. Showthat these observation are in agreement with the law of conservation of mass.Sodium carbonate + Ethanoic acid solution Sodium ethanoate Solution + Carbon dioxide
3
exam time—1
40
16. Calculate the molar mass of the following substances :(a) Ethyne (C2H2)(b) Sulphur molecule, S8(c) Phosphorus molecule, P4 (atomic mass of phosphorus = 31)(d) Hydrochloric acid, HCl(e) Nitric acid, HNO3. 5.
17. A silver ornament of mass ‘m’ gram is polished with gold equivalent to 1% of the massof silver. Compute the ratio of the number of atoms of gold and silver in the ornament.
5
Time allowed : 1½ Hrs. AVERAGE Max. Marks : 40
1. Define atom chemical. 12. Write the formula of the compound formed by the ions Al3+ and SO4
2–. 13. Give an example of a polyatomic cation. 14. Write the difference between 2 Cl and Cl2. 15. What is mole ? 16. How many molecules are present in 1 mL of water ? 27. Calculate the number of molecules in 4 g of methane. (Atomic mass of C = 12 u, H = 1 u)
28. What is the fraction of the mass of water due to neutrons ? 29. What is an ion ? Write the symbol of calcium ion and aluminium ion. 2
10. The formula of carbon dioxide is CO2. What information do you get from this formula ?(Any four) 2
11. Calculate the molecular masses of the following compounds :(a) Acetic acid (CH3 COOH) (b) Ethanol C2H5OH(c) Carbon dioxide.[At mass : C = 12 u, H = 1 u, O = 16 u] 3
12. (a) Convert 22 g of CO2 into mole(b) Write the chemical formula of copper sulphate(c) Calculate formula unit mass of CaO[Atomic mass : Ca = 40 u, O = 16 u] 3
13. What are the postulates of Dalton’s atomic theory ? 314. (i) Calculate the number of atoms in one gram of Zinc (atomic mass of Zinc is 65.37).
Give the weight of 1 atom of Zinc.(ii) How many moles are contained in 5 g of hydrogen ? 3
15. How is a molecular formula of a compound written ? 316. State and explain law of conservation of mass. How does Dalton’s atomic theory explain
this law 517. (a) Give the formula of baking soda. Name the elements present in baking soda. 5
exam time—2
41
(b) Calculate the number of moles for 84 g nitrogen atom. [Atomic mass of N = 14 u](c) Find out the percentage composition of sulphuric acid (H2SO4)
[Atomic mass H = 1 u, S = 32 u, O = 16 u]
Time allowed : 1½ Hrs. DIFFICULT Max. Marks : 40
1. Define molecule. 12. Write the chemical formula of calcium phosphate. 13. Give two examples of polyatomic molecules of elements. 14. Write the difference between 2 O and O2. 15. How many moles does 11 g of carbondioxide make ? 16. Calculate the number of molecules of sulphur (S8) in 16 g of solid sulphur. 27. Calculate the mass of 1 molecule of oxygen. (Atomic mass of oxygen = 16) 28. Differentiate between molecular mass and formula unit mass. 29. Write the formula of the following compounds :
(a) Calcium sulphite (b) Sodium phosphate(c) Ammonium carbonate (d) Magnesium nitride 2
10. (a) Define the term mole.(b) Calculate number of : (i) atoms, (ii) molecules in 12 u grams of phosphorus.[Given atomic mass of P = 31 u, NA = 6.022 × 1023 mol–1] 2
11. Calculate the number of particles in each of the following :(i) 46 g of Na atoms (number from mass)(ii) 8 g O2 molecules (number of molecules form mass)(iii) 0.1 mole of carbon atoms (number from given moles) 3
12. What is the mass of :(a) 1 mole of nitrogen atoms ?(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27 u).(c) 10 moles of sodium sulphite (Na2 SO3) ? 3
13. What are the postulates of modified Dalton’s atomic theory ? 314. Calculate percentage composition of glucose (C6H12O6). 315. Calculate the mass percentage of oxygen in (a) HgO (b) K2Cr2O7 (c) Al2 (SO4)3. The atomic
mass (in u units) are Mg = 24, O = 16, K = 39, Cr = 52, Al = 27, S = 32. 316. State and explain law of constant proportions. How does Dalton’s atomic theory explain
this law ? 517. (a) Calculate the number of molecules in 50 g of CaCO3 [Atomic mass : Ca = 40 u, C = 12
u, O = 16 u] 5(b) Calculate the number of moles of 112 g of iron u. [At. mass : N = 14 u](c) Calculate the number of moles of 112 g of iron u. [Atomic mass of Fe = 56 u](d) Calculate the formula mass of the following compounds :
(i) Magnesium sulphide (ii) Sodium chloride.[Atomic mass : Mg = 12u, S = 32u, Cl = 35.5 u].
exam time—3