ncert solutions for class 11 chemistry chapter 1 some

NCERT solutions for class 11 Chemistry chapter 1 Some Basic Concepts of Chemistry

Question 1.1 Calculate the molar mass of the following:

(i) H 2 O

Answer :

The molar mass of Water is:


(ii) CO 2

Answer :

The molar mass of Carbon dioxide is:

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(iii) CH 4

Answer :

The molar mass of Methane is:

Question 1.2 Calculate the mass per cent of different elements present in sodium

sulphate (Na 2 SO 4 ).

Answer :

The different elements present in sodium sulphate are:

Sodium (Na), Sulphur (S), and Oxygen (O).

The molar mass of sodium

sulphate ;

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So, Mass percentage of an element in a compound is given by,

Therefore,

The mass per cent of Sodium (Na):

The mass per cent of Sulphur (S):

The mass per cent of Oxygen (O):

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Question 1.3 Determine the empirical formula of an oxide of iron, which has 69.9% iron

and 30.1% dioxygen by mass.

Answer :

Given there is an oxide of iron which has iron and dioxygen by mass:

Relative moles of iron in iron oxide:

Relative moles of oxygen in iron oxide:

The simplest molar ratio of iron to oxygen:

Therefore, the empirical formula of the iron oxide is .

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Question 1.4 Calculate the amount of carbon dioxide that could be produced when

(i) 1 mole of carbon is burnt in air.

Answer :

When carbon is burnt in the air:

The chemical equation for this reaction is:

Here 1 mole of carbon (solid) weighing 12g is burnt in 1 mole of Dioxygen (gas)

weighing 32g to produced 1 mole of carbon-dioxide (gas) weighing 44g .


(ii) 1 mole of carbon is burnt in 16 g of dioxygen.

Answer :

When carbon is burnt in 16 g of dioxygen:


Here, oxygen acts as a limiting reagent as only 16g of dioxygen is available.

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Hence, it will react with 0.5 mole of carbon to give 22g of carbon dioxide.


(iii) 2 moles of carbon are burnt in 16 g of dioxygen.

Answer :

When 2 moles of carbon is burnt in 16 g of dioxygen:


Here, oxygen acts as a limiting reagent as only 16g of dioxygen is available.

Thus, 16g of dioxygen will react with 0.5 mole of carbon to give 22g of carbon dioxide.

Question 1.5 Calculate the mass of sodium acetate (CH 3 COONa) required to make

500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g

mol –1 .

Answer :

molar aqueous solution would contain moles of dissolved

in of solvent.

So, we have to calculate for solution of .

Therefore the number of moles of sodium acetate in will be:

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Given Molar mass of sodium acetate:

So, the required mass of sodium acetate

Question 1.6 Calculate the concentration of nitric acid in moles per litre in a sample

which has a density, 1.41 g mL - 1 and the mass percent of nitric acid in it being 69%.

Answer :

Given the mass percentage of nitric acid is .

That means 69 grams of nitric acid are present in 100 grams of nitric acid solution.

The molar mass of nitric acid is .

So, the number of moles in 69g of Nitric acid:

and volume of 100g of the nitric acid solution:

Therefore, the concentration of Nitric acid in moles per litre is:

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Question 1.7 How much copper can be obtained from 100 g of copper sulphate

(CuSO 4 )?

Answer :

Given that 100g of Copper sulphate ;

1 mole of contains 1 mole of copper.

Molar Mass of Copper sulphate is:

Now, of copper sulphate contains of copper.

So, of copper sulphate will contain copper content:

Question 1.8 Determine the molecular formula of an oxide of iron, in which the mass

percent of iron and oxygen are 69.9 and 30.1, respectively. Given that the molar mass

of the oxide is .

Answer :

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Given that the mass percentage of iron is and the mass percentage of oxygen

is

The atomic mass of iron .

The atomic mass of oxygen .

So, the relative moles of iron in iron oxide will be:

And relative moles of oxygen in iron oxide will be:

Hence the simplest molar ratio:

or

Therefore, the empirical formula of iron oxide will be .

Now, calculating the molar mass of :

.

Hence it is matching with the given molar mass of the oxide.

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Question 1.9 Calculate the atomic mass (average) of chlorine using the following data:

% Natural Abundance Molar Mass 35 Cl 75.77 34.9689 37 Cl 24.23 36.9659

Answer :

To calculate the average atomic mass of chlorine:

Given the fractional natural abundance of with molar mass

is and that of with molar mass is .

Therefore we have,

Average Atomic mass of Chlorine:

Question 1.10 In three moles of ethane (C 2 H 6 ), calculate the following:

(i) Number of moles of carbon atoms.

Answer :

Given there are three moles of Ethane ;

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So, 1 mole of contains 2 moles of carbon atoms.

Therefore, 3 moles of contains 6 moles of carbon atoms.


(ii) Number of moles of hydrogen atoms.

Answer :


So, 1 mole of contains 6 moles of hydrogen atoms.

Therefore, 3 moles of contains 18 moles of hydrogen atoms.


(iii) Number of molecules of ethane.

Answer :


So, 1 mole of contains molecules of ethane.

Therefore, 3 moles of contains molecules of

ethane.

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Question 1.11 What is the concentration of sugar (C 12 H 22 O 11 ) in mol L –1 if its 20 g

are dissolved in enough water to make a final volume up to 2L?

Answer :

The molar mass of sugar is:

.

The number of moles of sugar in 20g of sugar will be:

and given the volume of the solution after dissolving enough water is 2L .

Question 1.12 If the density of methanol is 0.793 kg L –1 , what is its volume needed for

making 2.5 L of its 0.25 M solution?

Answer :

Given that the density of methanol is .

So, the number of moles present in the methanol per litre will be or the Molarity of the

solution will be :

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Now, to make 2.5L of its 0.25M solution:

We will apply the

formula:

.

Hence volume will be required for making 2.5L of methanol 0.25M

solution.

Question 1.13 Pressure is determined as force per unit area of the surface. The SI unit

of pressure, pascal is as shown below:

1 Pa = 1 Nm –2

If mass of air at sea level is 1034 g cm –2 , calculate the pressure in pascal.

Answer :

The pressure is as given is force per unit area of the surface.

Given to calculate the pressure exerted by the air on sea water, if the mass of air at sea

level is .

The force with which the air is exerting on the surface is:

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Now,

as

Therefore,

Question 1.14 What is the SI unit of mass? How is it defined?

Answer :

The SI unit of mass is Kilogram (Kg).

It is defined as the mass of platinum-iridium (Pt-Ir) cylinder that is stored in an air-tight

jar at the International Bureau of Weigh and Measures in France.

Question 1.15 Match the following prefixes with their multiples:

Prefixes Multiples

(i) micro 10 6

(ii) deca 10 9

(iii) mega 10 –6

(iv) giga 10 –15

(v) femto 10

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Answer :

Matched items are given in below table:

Prefixes Multiples

(i) micro

(ii) Deca

(iii) Mega

(iv) Giga

(v) femto

Question 1.16 What do you mean by significant figures?

Answer :

Significant figures are meaningful digits that are known with certainty including the last

digit whose value is uncertain.

For example: if we write a result as 56.923 Kg, we say the 56.92 is certain and 3 is

uncertain and the uncertainty would be in the last digit. Here we also include the last

uncertain digit in the significant figures.

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Question 1.17 A sample of drinking water was found to be severely contaminated with

chloroform, CHCl 3 , supposed to be carcinogenic in nature. The level of contamination

was 15 ppm (by mass).

(i) Express this in per cent by mass.

Answer :

1ppm means 1 parts in million parts.

So, in percentage by mass:

We have,

Question 1.17 A sample of drinking water was found to be severely contaminated with

chloroform, CHCl 3 , supposed to be carcinogenic in nature. The level of contamination

was 15 ppm (by mass).

(ii) Determine the molality of chloroform in the water sample .

Answer :

The molarity of chloroform sample in water will be:

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The molar mass of Chloroform :

So we have calculated in the previous part that percentage by mass of chloroform

is .

Hence in 100g sample, there will be of chloroform.

Therefore, 1000g (1Kg) of the sample will contain choroform .

Therefore, the molarity of chloroform in the water sample is .

Question 1.18 Express the following in the scientific notation:

(i) 0.0048

Answer :

The scientific notation of will be .


(ii) 234,000

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Answer :



(iii) 8008

Answer :



(iv) 500.0

Answer :



(v) 6.0012

Answer :

The scientific notation of will be ???????.

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Question 1.19 How many significant figures are present in the following?

(i) 0.0025

Answer :

There are 2 significant digits because all non-zero digits are in a number are

significant and the zeros written to the left of the first non-zero digit in a number are non-

significant.


(ii) 208

Answer :


significant and the zeros between two non-zero digits are always significant.


(iii) 5005

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significant and the zeros between two non-zero digits are always significant.


(iv) 126,000

Answer :


significant and the terminal zeros are not significant if there is no decimal point.


(v) 500.0

Answer :

There are 4 significant digits because the zeros written to the left of the first non-zero

digit in a number are non-significant and all zeros placed to the right of a decimal point

in a number are significant.


(vi) 2.0034

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Answer :

There are 5 significant digits because all zeros placed to the right of a decimal point in

a number are significant.

Question 1.20 Round up the following upto three significant figures:

(i) 34.216

Answer :

After round upto three significant figures:

Answer -


(ii) 10.4107

Answer :


Answer - .


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(iii) 0.04597

Answer :

After round up to three significant figures:

Here the rightmost digit to be removed is more than 5 i.e., 7, then the preceding number

is increased by one. So we get .

Answer -


(iv) 2808

Answer :


Here the rightmost digit to be removed is more than 5 i.e., 8, then the preceding number

is increased by one. So we get .

Answer -

Question 1.21 The following data are obtained when dinitrogen and dioxygen react

together to form different compounds:

Mass of dinitrogen Mass of dioxygen

(i) 14 g 16 g

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(ii) 14 g 32 g

(iii) 28 g 32 g

(iv) 28 g 80 g

(a) Which law of chemical combination is obeyed by the above experimental data? Give

its statement.

Answer :

Here if we fix the mass of dinitrogen at 14g, then the masses of dioxygen that will

combine with the fixed mass of dinitrogen are 16g, 32g, 32g, and 80g.

The masses of dioxygen bear a whole number ratio of 1:2:2:5.

Hence, the given experimental data obeys the Law of Multiple Proportionals .

The law given by Dalton states that "if two elements can combine to form more than one

compound, the masses of one element that combine with a fixed mass of the other

element, are in the ratio of small whole numbers".

Question 1.21 The following data are obtained when dinitrogen and dioxygen react

together to form different compounds:

Mass of dinitrogen Mass of dioxygen

(i) 14 g 16 g

(ii) 14 g 32 g

(iii) 28 g 32 g

(iv) 28 g 80 g

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(b) Fill in the blanks in the following conversions:

(i) 1 km = ...................... mm = ...................... pm

(ii) 1 mg = ...................... kg = ...................... ng

(iii) 1 mL = ...................... L = ...................... dm 3

Answer :

(i) As we know in or .

or

And or

Therefore we have,

(ii) As we know in or .

or or

And or or

or .

Therefore we have,

.

(iii) As we know in or

or or

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And or we can write it as:

And

Therefore we have,

Question 1.22 If the speed of light is , calculate the distance covered

by light in 2.00 ns.

Answer :

Given the speed of light to be , so the distance covered by light in 2.00

ns. will be:

Therefore, the light will travel 0.600 metres in 2 nano seconds.

Question 1.23 In a reaction

Identify the limiting reagent, if any, in the following reaction mixtures.

(i) 300 atoms of A + 200 molecules of B

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Answer :

Here according to the reaction 1 atom of A reacts with 1 molecule of B.

Therefore, 200 molecules of atoms of B will react with 200 atoms of A, thereby left with

100 atoms of A unreacted.

Hence, B is the limiting reagent in this reaction.



(ii) 2 mol A + 3 mol B

Answer :


Therefore, 2 mol of A atoms will react with only 2 mol of B molecules, thereby left with 1

mole of B unreacted.

Hence, A is the limiting reagent in this reaction.



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(iii) 100 atoms of A + 100 molecules of B

Answer :


Therefore, all 100 atoms of A will react with 100 molecules of B, so the reaction is

stoichiometric and there is no limiting reagent .

Hence, there is no limiting reagent in this reaction.



(iv) 5 mol A + 2.5 mol B

Answer :


Therefore, 2.5 moles of B molecules will react with only 2.5 moles of A atoms, thereby

left with 2.5 moles of A unreacted.

Hence, B limiting reagent in this reaction.

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(v) 2.5 mol A + 5 mol B

Answer :


Therefore, 2.5 moles of A atoms will react with only 2.5 moles of B molecules, thereby

left with 2.5 moles of B unreacted.

Hence, A limiting reagent in this reaction.

Question 1.24 Dinitrogen and dihydrogen react with each other to produce ammonia

according to the following chemical equation:

(i) Calculate the mass of ammonia produced if 2.00 × 10 3 g dinitrogen reacts with 1.00

×10 3 g of dihydrogen.

Answer :

Given that if of dinitrogen reacts with of dihydrogen.

From the reaction we have:

1 mole of dinitrogen weighing 28g reacts with 3 moles of dihydrogen weighing 6g to give

2 moles of ammonia weighing 34g.

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Therefore, of will react with .

Thus, here is the limiting reagent while is in excess.

So, of produces of .

Therefore, of will produce of .



(ii) Will any of the two reactants remain unreacted?

Answer :

As from the previous part we have:

is the limiting reagent and is the excess reagent.

Hence, will remain unreacted.



(iii) If yes, which one and what would be its mass?

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Answer :

Yes , will remain unreacted.

And the mass of dihydrogen left unreacted will be .

Question 1.25 How are 0.50 mol Na 2 CO 3 and 0.50 M Na 2 CO 3 different?

Answer :

Calculating the molar mass of :

Therefore,

of means:

whereas,

means:

of or of are present in 1litre of the solution.

Question 1.26 If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen

gas, how many volumes of water vapour would be produced?

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Answer :

For the given situation we have the reaction:

Here, 2 volumes of dihydrogen gas react with 1 volume of dioxygen to produce 2

volumes of water vapour.

So, if 10 volumes of dihydrogen gas react with 5 volume of dioxygen then it will

produce ( volumes of water vapour.

Question 1.27 Convert the following into basic units:

(i) 28.7 pm

Answer :

To convert 28.7 pm into the basic units:

As .


(ii) 15.15 pm

Answer :

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To convert 15.15 pm into the basic units:

As .


(iii) 25365 mg

Answer :

To convert 25365 mg into basic unit:

As .

Now, as

Question 1.28 Which one of the following will have the largest number of atoms?

(i) 1 g Au (s)

(ii) 1 g Na (s)

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(iii) 1 g Li (s)

(iv) 1 g of Cl 2 (g)

Answer :

Calculating and then comparing for each:

(i) 1 g of Au will contain:

(ii) 1 g of Na will contain:

(iii) 1 g of Li will contain:

(iv) 1 g of Cl 2 will contain:

Clearly, we can compare and say that the number of atoms in 1g of Li has the

largest.

Question 1.29 Calculate the molarity of a solution of ethanol in water, in which the mole

fraction of ethanol is 0.040 (assume the density of water to be one) .

Answer :

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Given that the mole fraction of ethanol in water is .

To find the molarity we must have to find the number of moles of ethanol present in

1litre of solution.

Assuming the density of water to be .

Therefore, water is approximately equal to 1Litre.

The number of moles in 1L of water:

So, substituting in place of in above equation we get,

So, are present in 1L of solution.

Hence, the molarity of the solution is

Question 1.30 What will be the mass of one 12 C atom in g?

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Answer :

As we know the mass of 1 mole of atoms or number of atoms

is 12grams . Hence, the mass of one atom will

be:

Question 1.31 How many significant figures should be present in the answer of the

following calculation?

(i)

Answer :

To find the number of significant figures that would be present in the answer, we will be

finding theleast precise term, having the least significant figures.

Here, the least precise term is 0.112 having only 3 significant digits.

Therefore, there will be 3 significant figures in the calculated answer.



(ii)

Answer :

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Here, is having 4 significant digits.

Therefore, after multiplying by 5 the answer would also have the same 4 significant

figures .



(iii) 0.0125 + 0.7864 + 0.0215

Answer :

Here, the least number of decimal places in each term is four.

Therefore, the calculation would also have the same 4 significant figures .

Question 1.32 Use the data given in the following table to calculate the molar mass of

naturally occuring argon isotopes:

Isotope Isotopic molar mass Abundance

36 Ar 35.96755 g mol -1 0.337%

38 Ar 37.96272 g mol -1 0.063%

40 Ar 39.9624 g mol -1 99.600%

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Answer :

For different isotopes of argon, we have given their naturally occurring abundances.

So, to calculate the molar mass:

Multiply the isotopic molar mass with their abundance to get the molar mass, and then

add all of them to get,

Question 1.33 Calculate the number of atoms

(i) 52 moles of Ar

Answer :

As 1 mole of air contains atoms.

Therefore, 52 moles of Ar will contain atoms.

.

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(ii) 52 u of He.

Answer :

As 1 atom of He weights 4 u.

Hence the number of atoms of He present in 52 u will be:


(iii) 52 g of He.

Answer :

As 1 mole of He weights 4g and contains atoms.

Therefore, will contain .

Hence, the number of atoms will be .

.

Question 1.34 A welding fuel gas contains carbon and hydrogen only. Burning a small

sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other

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products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh

11.6 g. Calculate

(i) empirical formula

Answer :

The amount of carbon in of :

The amount of hydrogen in of :

The compound contains only C and H,

Therefore, the total mass of the compound will be:

Now, the percentage of Carbon in the compound:

and the percentage of Hydrogen in the compound:

Now, the empirical formula,

Moles of carbon in the compound:

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Moles of hydrogen in the compound:

So, the simplest molar ratio will be

Therefore, the empirical formula is .




11.6 g. Calculate

(ii) molar mass of the gas

Answer :

Wight of 10 Litres of the gas at S.T.P. is

Weight of 22.4 Litres of gas at S.T.P. will be:

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11.6 g. Calculate

(iii) molecular formula.

Answer :

As we know the empirical formula from the previous part is .

The mass of empirical formula .

Therefore,

Therefore, Molecular formula is .

Question 1.35 Calcium carbonate reacts with aqueous HCl to give CaCl 2 and

CO 2 according to the reaction,

What mass of CaCO 3 is required to react completely with 25 mL of 0.75 M HCl?

Answer :

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contains in of solution.

Or, of in solution.

Therefore,

Mass of in of :

so, from the given chemical equation,

1 mole of i.e., reacts with 2 moles of i.e., .

Therefore, reacts completely with to give:

Question 1.36 Chlorine is prepared in the laboratory by treating manganese dioxide

(MnO 2 ) with aqueous hydrochloric acid according to the reaction

How many grams of HCl react with 5.0 g of manganese dioxide?

Answer :

Molar mass of is .

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Here from the reaction 1 mole of reacts with 4 moles of ,

i.e., of .

Therefore, of will react with :

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ncert solutions for class 11 chemistry chapter 1 some

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