152220446 chapter 9 solutions text book exercise

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CHAPTER 9

SOLUTIONS

TEXT BOOK EXERCISE Q1. Choose the correct answer for the given ones. (i) Morality of pure water is

(a) 1. (b) 18. (c) 55.5 (d) 6.

Hint: Morality of pure water

Consider 1 dm3 (-1000cm

3 ) of water. Convert this volume into mass by

using density .

Mass =volume x density

Mass of H2O = 1000 cm3 x 1 g cm

3 =1000g

Molar mass of H2O =18 g mol-1

No. of moles of H2O = =55.6 mol

Morality of H2O =

= =55.6 mol dm-3

=55.6 M

(ii) 18 g glucose is dissolved in 90 g of water. The relative lowering

of vapour pressure is equal to

(a) 1 (b) 18 (c) 55.5 (d) 6

(iii) A solution of glucose is 10%. The volume in which 1 g-mole of it

is dissolved will be

(a) 1 dm3 (b) 1.8dm

3 (c) 200 cm

3 (d)

900cm3

(iv) An aqueous solution of ethanol in water may have vapour

pressure

(a) equal to that of water (b) equal to that of

ethanol

(c) more than that of water (d) less than that of

water


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(v) An azeotropic mixture of two liquids boils at a lower temperature

than either of them when

(a) it is saturated

(b) it shows positive deviation from Raoults law (c) it shows negative deviation from Raoults law (d) it is met stable

(vi) In azeotropic mixture showing positive deviation from Raoults law , the volume of the mixture is

(a) slightly more than the total volume of the components

(b) slightly less than the total volume of the components

(c) equal to the total volume of the components

(d) none of these

(vii) Which of the following solution has the higher boiling point?

(a) 5.85% solution of sodium chloride

(b) 18.0% solution of glucose

(c) 6.0% solution of urea

(d) All have the same boiling point.

(e) Two solution of NaC1and KC1 are prepared separately by

dissolving same amount of solute in water. Which of the following

statements is true for these solutions?

(f) KC1 solution will have higher boiling point than NaC1

solution

(g) Both the solution have different boiling points.

(h) KC1 and NaC1 solutions possess same vapour pressure

(i) KC1 solution possesses is the ratio of the elevation in

boiling point to

(j) The molal boiling point constant is the ratio of the elevation

in boiling point to

(a) molaritly (b) molality

(c) mole fraction of solvent (d) mole fraction of solute

(x) Colligative properties are the properties of

(a) Dilute solutions which behave as nearly ideal non-ideal

solutions


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(b) Concentrated solutions which behave as nearly non-ideal

solutions

(c) Both (a) and (b)

(d) Neither (a) and (b)

Ans. (i) c (ii) c (iii) b (iv) c

(v) b

(vi) a (vii) a (viii) c (ix) b

(x) a

Q2. Fill in the blanks with suitable words or numbers. (i) Number of molecules of sugar in 1dm

3 of 1 M sugar solution

is _________.

(ii) 100 g of a 10% aqueous solution of NaOH contains 10 g

NaOH in _________ g of water.

(iii) When an azeotropic mixture is distilled, its

__________remains constant.

(iv) The molal freezing point of an azeotropic solution of two

liquids is lower than ether of them because the solution

shows________from Raolulss law. (v) Among equimolal aqueous solution of NaC1, BaC12, and

FeC13, the maximum depression in freezing point is shown by

_______solution.

(vi) A solution of ethanol in water shows____deviations and gives

azeotropic solutions with _______b.p. than other components.

(vii) Colligative properties are used to calculate _______of a

compound.

(viii) The boiling point of an azeotropic solution of two liquids is

lower than either of them because the solutions show

________from Raoults law. (ix) The hydration energy of Br

- ion is _________than that of F

-

ion.

(x) The aqueous solution of NH4C1 is _________while that of

Na2SO4 is _______.


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Ans. (i)6.02x 1023

(ii)90

(iii)composition (iv)cryoscopic (v)Positive deviation

(vi)FeC13 (vii)positive ; lower

(viii) molar mass (ix) lesser (x) acidic ;neutral

Q3. Indicate True or False from the given statements. (i) At a definite temperature the amount of a solute in a

given saturated solution is fixed.

(ii) Polar solvents readily dissolve non-polar covalent

compounds.

(iii) The solubility of a substance decreases with increase in

temperature if the heat of a solution is negative.

(iv) The rate of evaporation of a liquid is inversely proportional to

the intermolecular forces of attraction.

(v) The molecular mass of an electrolyte determined by lowering

of vapour pressure is less than the theoretical molecular mass.

(vi) Boiling point elevation is directly proportional to the molality

of the solution and inversely proportional to boiling point of

solvent.

(vii) All solutions containing 1 g of non-volatile non-electrolyte

solutes in some solvent will have the same freezing point.

(viii) The freezing point of a 0.05 molal solution of a non-volatile

non-electrolyte solute in water is 0.93oC.

(ix) Hydration and hydrolysis are different process for Na2SO4.

(x) The hydration energy of an ion only depends upon its charge.

Ans. (i) True (ii)False (iii)True (iv)True

(v)True

(vi) True (vii) False (viii) False (ix)True (x)

False

Q4. Define and explain the following with examples: (a) A homogeneous phase (b) A concentrated solution

(c) A solution of solid in a solid (d) A consulate temperature

(e) A non-ideal solution (f) Zeotropic solutions

(g) Heat of hydration (h) Water of crystallization


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(i) Azeotropic solution (j) Conjugate solution

Q5. (a) What are the concentration units of solutions? Compare

molar and molal solution.

Ans. Compare molar and molal solution.

Molar Solution Molar Solution

1. A solution which contains one mole of solute per dm

3 of

solution is called a molar

solution. 2. Its morality decreases with

the rise in temperature.

3. Its unit is mole dm3 of

solution.

1. A solution which contains one mole of solute per kilogram

of solvent is called a molal

solution. 2. Its molality is independent of

temperature variation of

solution.

3. Its unit is mol kg-1

of

solvent.

(b) One has one molal solution of NaC1 and one molal solution of

glucose.

(i) Which solution has greater number of particles of solute?

(ii) Which solution has greater amount of the solvent.?

(iii) How do we convert these concentrations into weight by

weight percentage?

(b) (i) One molal solution of NaC1 has greater number of

particles of solute.

(ii) The amount of solvent in both solution is equal (1 kg)

(iii) Mass of NaC1 = 58.5 g

Mass of solvent =1000g

Total mass of NaC1 solution =1000 + 58.5 =1058.5 g

Percentage of NaC1 = x100

= x100=5.53 % Answer

Mass of glucose =180g

Mass of solvent =1000g


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Total mass of solution =1000+180=1180 g

% of glucose = x100

= x100=15.25 % Answer

Q.6 Explain the following with reasons (i) The concentrations in terms of molality is independent of the

temperature but morality depends upon temperature.

Ans. Molality is based on the mass of solvent. The mass of solvent does

not vary not vary with temperature, so molality is independent of

temperature. Morality is based on the volume of solution. Since the

volume of solution varies with temperature, so morality depends upon

temperature. The morality of the solution decreases with the increase in

temperature of the solution.

(ii) The sum of the mole fractions of all the components is always

equal to unity for any solution.

Ans. Suppose there be two components A and B making a solution.

The numbers of moles are nA and nB respectively, then. If the mole

fractions of A and B are denoted by XA and XB respectively, then

XA = and XB =

The sum of mole fractions of the components of a solution will be

XA + XB = +

XA + XB =

XA + XB =1

Hence, the sum of the mole fractions of all the components of any

solution is always equal to unity.

(iii) 100 g of 98% H2SO4 has a volume of 54.34 cm3 of H2SO4 since

its density is 1.84 g cm-3

.

Ans. Mass of 98% H2SO4 =100g : density of 98% =1.84 g cm-3

Vol. of 98% H2SO4=?


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Formula Used: volume = =

=54.34 cm-3

Hence, 100 g of 98 % H2SO4 has a volume of 54.34 cm-3

of

H2SO4 because its density is 1.84 g cm-3

.

(iv) Relative lowering of vapour pressure is independent of the

temperature.

Ans. Since, the relative lowering of vapour pressure is equal to the

mole fraction of solute, so it is independent of temperature.

Hint: =X2 or =

(v) Colligative properties are obeyed when the solute is non-

electrolyte and also when the solutions are dilute.

Ans. Colligative properties are obeyed when the solute is non-

electrolyte

Colligative properties depend only upon the number of solute

particles and not on their chemical nature. An electrolyte solute differs

from a non-electrolyte principally in the number of particles produced

upon dissolution. In case the solute is non-electrolyte , one mole of solute

produces one mole of dissolved particles (molecules). In case the solute is

electrolyte, it may split into a number of ions each of which acts as a

particle and thus will affect the colligative properties of solution than

non-electrolytes. For example, one mole of glucose produces one mole of

dissolved particles (molecules) while one mole of NaC1 produces two

moles of dissolved particles (one mole each of Na+ and C1

- ions). Thus,

mole for mole, the NaC1 exerts twice the colligative effect than glucose

if the solution is ideal.

Colligative properties are obeyed when the solutions are

dilute. Colligative properties are obeyed when the solutions are dilute. A

dilute solution behaves almost as an ideal solution, i.e., the solute-solute

interactions are negligible. Concentrated solution is mostly non-ideal.

(vi) The total volume of the solution by mixing 100 cm3 of

water with 100 cm3 of alcohol may not be equal to 200 cm

3. Justify it.


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Ans. Because the intermolecular forces of attraction between alcohol

and water molecules are not the same as the intermolecular attractive

forces between alcohol molecules or between water molecules. Hence,

the total volume of the solution by mixing 100 cm3 of alcohol with 100

cm3 of water will not be equal to 200 cm

3. The total volume of solution

will be greater than 200 cm3 because the forces of attraction between

alcohol and water molecules are weaker than those between alcohol

molecules or between water molecules.

(vii) One molal solution of urea in water is dilute as compared

to one molar solution of urea, but the number of particles of the solute is

the same. Justify it.

Ans. One molal solution of urea in water is dilute as compared to one

molar solution of urea. This is because a molal solution contains one

mole of urea in 1000 g of water whereas one molar solution contains one

mole of urea in 1000 cm3 of water.

At room temperature, density of water is slightly less than one.

Therefore, the volume corresponding to 1000 g of water be greater than

1000cm3. So, the volume of solvent water containing one mole of solute

is more in case of molal solution than molar solution. hence, one molal

solution of urea in water is dilute as compared to one molar solution of

urea in water. Since both the solutions contain 1 mole of urea as solute,

therefore , the number of particles of solute is the same in both the

solution.

(viii) Non-ideal solution do not obey the Raoults law. Ans. They show deviations from Raoults law due to differences in their molecular structures, i.e. , size , shape and intermolecular forces.

Formation of such solutions is accompanied by changes in volume and

enthalpy. The vapour pressure deviation may be positive or negative in

such solutions.

Q7. What are non-ideal solutions? Discuss their types and give three

example of each.

Q8. (a) Explain fractional distillation. Justify the two curves when

composition is plotted against boiling point of solutions.


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(b) The solution showing positive and negative deviations

cannot be fractionally distilled at their specific compositions. Explain it.

Q9. (a) What are azeotropic mixtures? Explain them with the help

of graphs.

(b) Explain the effect of temperature on phenol-water system.

Q10. (a) What are collligative properties? Why are they called so?

Ans. Because the colligative properties of solution depend only upon

the number of solute and solvent particles present in the solution and not

upon the chemical nature of the solute molecules. For this reason these

properties are called colligative properties.

(b) What is the physical significance of KB and kf values of

solvent?

Ans. Because the change in boiling point and the freezing point of a

solvent is a colligative property that depends only on the ratio of the

number of particles of solute an dissolvent in the solution, so these

constants are used to determine the molecular mass of an unknown

solute.

Q11. How to explain that the lowering of vapour pressure is a

colligative property?

How do we measure the molar mass of a non-volatile, non-

electrolyte solute in a volatile solvent?

Q12. How do you justify that?

(a) Boiling points of the solvents increase due to the presence

of solutes.

Ans. The temperature of a pure liquid at which its vapour pressure becomes equal to an external atmospheric pressure on its surface is

called boiling point of the liquid. The addition of a non-volatile solute to a solvent always reduces

its vapour pressure below that of the pure solvent. This is because solute

particles occupy some of the surface area of the solution decreases the

rate of evaporation and thus reduces its pressure. So, a higher temperature

is needed to increase the vapour pressure to the point where the solution


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boils. Hence, the boiling points of the solvents increase due to the

presence of solutes.

(b) Freezing points are depressed due to the presence of

solutes.

Ans. The temperature f a pure liquid at which its solid and liquid are froms coexist in equilibrium is called freezing point of the liquid. When a non-volatile solute is added to a solvent, it reduces its

vapour pressure. If a solution is cooled sufficiently, the temperature at

which crystals of pure solvent appear is the freezing point of the solution.

At this temperature the solid solvent and solution are in equilibrium, so

they must have the same vapour pressure. But , at a definite temperature,

a solution containing a non-volatile solute has a lower vapour pressure

than the pure solvent. Therefore, solid solvent must be in equilibrium

with a solution at a lower temperature than the temperature that it would

be in equilibrium with pure solvent. Hence, freezing points are depressed

due to the presence of solutes.

(c) The boiling point of one molal urea solution is 100 .52o

C

but the boiling of two molal urea solution is less than 101.04oC.

Ans. The boiling point of 1 molal urea solution is 100.52oC at 1 atm

pressure rather than 100oC. Thus the elevation of boiling point is (100.52

100.00) =0.52oC. The boiling point of 2 molal urea solution is 101.04

oC , so the elevation of boiling points is (101.04 100.00)= 1.04

oC. Since

the boiling point elevation depends upon the number of particles of

solute, therefore, e molal urea solution which contains 2 x 6.02 x 1023

molecules has double the boiling point elevation than 1 molal urea

solution which contains 6.02 x1023

molecules.

(d) Beckmanns thermometer is use to note the depression in freezing point.

Ans. Beckmanns thermometer is used to read temperatures up to, 0.01

oC over a range of about 5

oC. Since, freezing point depressions are


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small , no more than a degree or two. Therefore, to measure a small

difference in temperature Beckmanns thermometer is used. It is more sensitive than ordinary thermometer because one degree is further

divided into hundred divisions.

(e) In summar the antifreeze solutions protect the radiator

from boiling over.

Ans. Water is used as a coolant in automobile radiation to decrease the

temperature of the working engine. Pure water boils at 100oC. It is

observed that solutions boil at higher temperatures than do pure liquid.

So, an aqueous solution of antifreeze such as ethylene glycol is used in

place of pure water in radiators because it raises the boiling point of pure

water. Hence, in summer the antifreeze solution protects the liquid of the

radiator to boil over.

(f) NaC1 and KNO3 are used to lower the melting point of

ice.

Ans. It is a common observation that the freezing point of solution is

always lower than the freezing point of the pure solvent. The lowering in

freezing point depends upon the number of solute particles (molecules/

ions).

A mixture of NaC1 and KNO3 salts is used as a freezing mixture

to lower melting points of ice. These salts dissociate in ice water. They

split up into a number of ions each of which as a particle due to which the

freezing point of water, i.e., the melting point of ice is lowered.

Q13. What is Raoults law? Give is three statements. How this law can help us to understand the ideality of a solution?

Q14. Give graphical explanation for elevation of boiling point of a

solution. Describe one method to determine the boiling pointy elevation

of a solution.

Q15. Freezing points of solutions are depressed when non-volatile

solutes are present in volatile solvents. Justify it. Plot a graph to elaborate


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your answer. Also give one method to record the depression of freezing

point of a solution.

Q16. Discuss the energetics of solution. Justify the heats of solutions as

exothermic and endothermic properties.

Q17. (a) Calculate the molarity of glucose solution when 9 g of it

are dissolved in 250 cm3 of solution.

Solution: (a) Mass of glucose C6H12O6 =9g

Molar mass of C6H12O6 =180g mol-1

Vol. of solution =250 cm3

=0.25dm3

Molarity =?

Formula Used: Molarity= x

Molarity = x

Molarity =0.2 mol dm-3

Answer

(b) Calculate the mass of urea in 100 g of H2O in 0.3 molal solutions.

Solution: Molality=0.3

Mass of H2O =100 g =0.1 kg

Mass of urea (solute) =?

Molar mass of urea, NH2CONH2=60 g mol-1

Formula Used: Molarity= x

0.3 mol kg-1

= x

Mass of urea = 0.3 mol kg-1

x 60 g mol-1

x 0.1 kg

Mass of urea = 1.8 g Answer

(c) Calculate the concentration of a solution in moles kg-1

, which is

obtained by mixing 250 g of 20% solution of NaC1 with 200 g of 40%

solution of NaC1.

Solution: First solution:


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100 g of NaC1 solution contains pure NaC1 =20g

250 g of NaC1 solution contains pure NaC1 = x250 g =80g

Second solution:

100 g of solution contains pure NaC1 =40g

200 g of NaC1 solution contains pure = x200 g =80g

Total mass of solute NaC1 =(50 + 80)g =130 g

Total mass of NaC1 solution =(250 + 200) g =450 g

Mass of solvent =(450 130)g =320 g =0.32 kg Formula used:

Molarity = x

Molality = x

Molality=6.94 mol kg-1

Answer

Q18. (a) An aqueous solution of sucrose has been labeled as 1

molal. Find the mole fraction of the solute and the solvent.

Solution: Molality =1

Mass of solute, C12H22O11 =?

Molar mass of C12H22O11 =342 g mol-1

No. of kg of solvent =1 kg

Formula used:

Molarity = x

Mass of solute, =molality x molar mass of solute x No. of Kg of solvent

= 1 x 342 x 1

=342 g

Now, No. of moles of sucrose= =1mol

No. of moles of water = =55.56 mol

Total no. of moles =1 + 55.56 = 56.56 moles


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Mole fraction of sucrose = =0.076 Answer

Mole fraction of water = =0.9823 Answer

(b) Your are provide with 80% H2SO4 having density 1.84 g cm-3.

How much volume of this H2SO4 sample is required to obtain one dm3 of

20% H2SO4 which has a density of 1.25 g cm-3

.

Solution: First solution: % of H2SO4 =80

Density of H2SO4 = 1.8 g cm-3

It means that: 1 cm3 of H2SO4 has mass =1.8 g

1000 cm3 H2SO4 has mass =1.8 x1000=1800g

Now, 100g of H2SO4 solution contains pure H2SO4 =80 g

1800 g of H2SO4 solution contains pure H2SO4 = x 1800 =1440 g

No. of moles of H2SO4 =

= =14.7 mol

So, 14.7 moles of H2SO4 are present in 100 cm3 of concentrated H2SO4

solution, therefore , the molarity of solution is 14.7.

Molarity of conc. H2SO4 = 14.7 M

Second solution: Percentage of dilute H2SO4 solution =20% (w/w)

Density of dilute H2SO4 solution =1.25 g cm-3

It means that: 1 cm3 of H2SO4 has mass=1.25 g

1000 cm3 of H2SO4 has mass =1.25 x 1000 =1250 g

Now, 100 g dilute H2SO4 solution contains pure H2SO4 =20g

1250 g of dilute H2SO4 solution contains pure H2SO4 = x1250 =250 g

No. of moles of H2SO4 = =2.55 mol

So, 2.55 moles of H2SO4 are present in 1000 cm3 of dilute H2SO4

solution, therefore the molarity of dilute H2SO4 solution is 2.55 M.

Now, volume of conc. H2SO4 solution required to prepare dilute H2SO4

solution can be calculated by using the dilution formula:


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Conc. H2SO4 dilute H2SO4

M1V1 = M2V2

14.7 xV1 = 2.55 x 1000cm3

V1

V1=173.46 cm3 =173.5 cm

3

Hence, volume of concentration H2SO4solution required to prepare dilute

H2SO4

Solution =175.5 cm3 Answer

Q19. 250 cm3

of 0.2 molar K2SO4 solution is mixed with 250 cm3

of

0.2 molar KC1 solution. Calculate the molar concentration of K+ ions in

the solution.

Solution:

K2SO4 2K++ SO

0.2 M 2 x 0.2

0.4

Molarity of K+ ions=0.4M

KC1 K++ C1

0.2 M 0.2M

Molarity of K+

ions =0.2M

Total molarity of K+ ions=(0.4 +0.2) M=0.6M

Total volume of solution =250 cm3 +250 cm

3=500cm

3

Since after mixing the two solutions, the total volume becomes 500 cm3,

so the concentration of K+

ions becomes half. So,

Molarity of K+

ions = =0.3 M Answer

Q20. 5 g of NaC1 are dissolved in 1000 g of water. The density of

resulting solution is 0.997 g cm-3

. calculate molality, molarity and mole-

fraction of this solution.

Assume that the volume of the solution is equal to that of solvent.

Solution:

(i) Calculations for Molality Mass of solute, NaC1 =5g

Molar mass of solute =58.5 g mol-1


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Mass of solvent , H2O =1000g =1kg

Molality=?

Formula used:

Molarity = x

Molality =

=0.0854 mol kg-1

= 0.0854 m Answer

(ii) Calculations for molarity: Mass of solute, NaC1 =5g

Molar mass of solute =58.5 g mol-1

Mass of solvent, H2O =1000 g = 1kg

Density of solution=0.997 g cm3

Now, d or V=

Vol. of solution = =1003 cm3

Vol. of solution in dm3 = =1.003 cm

3

Molarity = x

=

=0.0852 mol dm-3

= 0.0852 M Answer

(ii) Calculations for mole fraction:

No. of moles of solute, NaC1 = =0.0855 mol

No. of moles of solvent H2O = =55.556 mol

Total number of moles =0.0855 + 55.556 =55.64 mol

Mole fraction of NaC1, X NaC1= =0.00154 Answer

Mole fraction of water, XH2O = =09984 Answer

Q21. 4.675 g of compound with empirical formula C3H3O were

dissolved in 212.5 g of pure benzene. The freezing point of solution was


17


found 1.02oC less than that of pure benzene. The molal freezing point

constant of benzene is 5.1oC. calculate (i) the relative molar mass and (ii)

the molecular formula of the compound.

Solution: (i) Mass of solute=4.675 g

Mass of solvent =212.5 g

Kf =5.1oC

Tf =1.02oC

Molar mass of solute =?

Formula used:

Molar mass of solute = = x 1000

Molar mass of solute = =

Molar mass of solute =110 g mol-1

Answer

(ii) Empirical formula =C3 H3O

Empirical formula mass =36+3+16=55

n =

n = =2

Molecular formula =(Empirical formula)n

=( C3 H3O)2

= C3 H3O2 Answer

Q22. The boiling point of a solution containing 0.2 g of a substance A

in 20.0 g of ether (molar mass =74) is 0.17 K higher than that of pure

ether. Calculate the molar mass of A. Molal boiling point constant of

ether is 2.16 K.

Solution: Mass of A=0.2 g

Mass of solvent =20 g

Tb =0.17 k

Kb =2.16K

Molar mass of solute, A=?


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Formula used: Molar mass of A x x1000.

Molar mass of A = x x1000

Molar mass of A=127 g mol-1

Answer

Q23. 3 g of a non-volatile , non-electrolyte solute X are dissolved in 50 g of ether (molar mass =74) at 293 K. The vapour pressure of ether

falls from 442 torr to 426 torr under these conditions. Calculate.

Calculate the molar mass of solute X. Solution: Mass of solute, X =3g

Mass of solvent (ether) =50 g

Molar mass of solvent (ether) =74

Vapour pressure of pure solvent, Po =442 torr

Vapour pressure of solution, p =426 torr

P=Po p =442 426 =16 torr

Molar mass of solute, X =?

Formula used:

Molar mass of solute, X = x x Molar mass of solvent

=

=122.66 g mol-1

Answer

152220446 chapter 9 solutions text book exercise

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