Two liquids, A & B , form an ideal solution. At the specified temperature, the vapour pressure of pure A is 200 mm Hg while that of pure B is 75 mm Hg . If the vapour over the mixture consists of 50 mol % A , what is the mole per cent A in the liquid?

Given, in the vapour phase mole fraction of liquid A = 50 % = 50 100 = 0 . 5 Therefore, the mole fraction of liquid B = 1 - 0 . 5 = 0 . 5 Y A = P A X A P t Y B = P B X B P t Y A = Y B P A = 200 , P B = 75 P A X A P B = X B 200 X A 75 = 1 - X A 8 X A 3 = 1 - X A on solving X A = 3 11 = 0 . 27 in percent 0 . 27 × 100 = 27 %

Complete the following table. Solute Dissociation/association reaction Degree of dissociation/association n i KCl 1 H 2 SO 4 1 CH 3 COOH (in water) 0 . 2 CH 3 COOH (in benzene) 0 . 5 Urea NaBr 0 . 8 A 3 A → A 3 1

The relation between the degree of dissociation or association of the solute and the Van't Hoff factor i is, For dissociation: α = i - 1 n - 1 . For association: α = i - 1 1 n - 1 . Here, n is the number of ions of the electrolytes and α is the degree of dissociation or association. The electrolytes KCl , H 2 SO 4 , CH 3 COOH aqueous and NaBr show dissociation and their number of ions is equal to n . CH 3 COOH in benzene exists as a dimer due to hydrogen bonding. Urea NH 2 CONH 2 is a non-electrolyte. So, n = 1 and i = 1 . KCl H 2 SO 4 CH 3 COOH in water CH 3 COOH in benzene NaBr 3 A → A 3 α = i - 1 n - 1 α = i - 1 n - 1 α = i - 1 n - 1 α = i - 1 1 n - 1 α = i - 1 n - 1 α = i - 1 1 n - 1 1 = i - 1 2 - 1 1 = i - 1 3 - 1 0 . 2 = i - 1 2 - 1 0 . 5 = i - 1 1 2 - 1 0 . 8 = i - 1 2 - 1 1 = i - 1 1 3 - 1 i = 2 i = 3 i = 1 . 2 i = 0 . 75 i = 1 . 8 i = 0 . 33 Solute Dissociation/association reaction Degree of dissociation/association n i KCl KCl → K + + Cl - 1 2 2 H 2 SO 4 H 2 SO 4 → 2 H + + SO 4 2 - 1 3 3 CH 3 COOH (in water) CH 3 COOH → H + + CH 3 COO 0 . 2 2 1 . 2 CH 3 COOH (in benzene) 2 CH 3 COOH → CH 3 COOH 2 0 . 5 0 . 5 0 . 75 Urea No association or dissociation No association or dissociation Not defined 1 NaBr NaBr → Na + + Br 0 . 8 2 1 . 8 A 3 A → A 3 1 0 . 33 0 . 33

Choose the correct statement with respect to the vapour pressure of a liquid among the following :

Increases non-linearly with increasing temperature

Decreases non-linearly with increasing temperature

Decreases linearly with increasing temperature

Increases linearly with increasing temperature

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IMPORTANT

Earn 100

Two liquids $A & B$ form an ideal solution. At $300 K$ , the vapour pressure of a solution containing $1 mole of A$ and $3 moles of B$ is $550 mm of Hg$ . At the same temperature, if one mole of $B$ is added to this solution, the vapour pressure of the solution increases by $10 mm of Hg$ . Determine the vapour pressure of $A and B$ in their pure states.

Important Questions on Solutions

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 8 - Solutions>Exercise-1>Q 16

Two liquids,

A & B

, form an ideal solution. At the specified temperature, the vapour pressure of pure

A

200 mm Hg

while that of pure

B

75 mm Hg

. If the vapour over the mixture consists of

50 mol % A

, what is the mole per cent

A

in the liquid?

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 8 - Solutions>Exercise-1>Q 17

Two solutions of

A & B

are available. The first is known to contain

1 mole of A & 3 moles of B

and its total vapour pressure is

1.0 atm

. The second is known to contain

2 moles of A & 2 moles of B

its vapour pressure is greater than

1 atm

, but it is found that this total vapour pressure may be reduced to

1 atm

by the addition of

6 moles of C

. The vapour pressure of pure

C

0.80 atm

. Assuming ideal solutions and that all these data refer to

25^{0} C

, calculate the vapour pressure of pure

A

and of pure

B

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 8 - Solutions>Exercise-1>Q 18

80^{0} C

, the vapour pressure of pure benzene is

753 mm Hg

and of pure toluene

290 mm Hg

. Calculate the composition of a liquid in mole per cent at

80^{0} C

in equilibrium with the vapour containing.

30 mole %

of benzene.

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 8 - Solutions>Exercise-1>Q 19

The vapour pressure of

C_{6} H_{6}

and

C_{7} H_{8}

mixture at

50^{0} C

is given by

P (mm Hg) = 180 X_{B} + 90

, where

X_{B}

is the mole fraction of

C_{6} H_{6}

. A solution is prepared by mixing

12 mol C_{6} H_{6}

and

8 mol C_{7} H_{8}

and if vapours over this solution are removed and condensed into liquid and again brought to the temperature

50^{0} C

, what would be the mole fraction of

C_{6} H_{6}

in the vapour state? (Atomic weight of

C = 12, H = 1

)

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 8 - Solutions>Exercise-1>Q 20

A non-ideal solution was prepared by mixing

30 mL

chloroform and

50 mL

acetone. Comment on the volume of the mixture.

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 8 - Solutions>Exercise-1>Q 21

The total vapour pressure of a mixture of

1 mol

of volatile component

A ({P_{A}}^{0} = 100 mm Hg)

and

3 mol

of volatile component

B ({P_{B}}^{0} = 80 mm Hg)

90 mm Hg

. Find out the nature and the sign of the entropy of the solution.

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 8 - Solutions>Exercise-1>Q 22

Complete the following table.

Solute	Dissociation/association reaction	Degree of dissociation/association	n	i
$KCl$		$1$
$H_{2} {SO}_{4}$		$1$
${CH}_{3} COOH$ (in water)		$0.2$
${CH}_{3} COOH$ (in benzene)		$0.5$
Urea
$NaBr$		$0.8$
A	$3 A \to A_{3}$	$1$

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 8 - Solutions>Exercise-1>Q 23

Calculate the percentage of the degree of dissociation of an electrolyte

{XY}_{2}

(Normal molar mass

= 164

) in water if the observed molar mass by measuring the elevation in boiling point is

65.6

Important Questions on Solutions

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