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CHXII02:SOLUTIONS

319480 Which one of the following pairs will show positive deviation from Raoult's Law?

1 Water - HCl

2 Benzene-Methanol

3 Water

- {HNO}_{3}

4 Acetone - Chloroform

Explanation:

Benzene $- {CH}_{3} OH$ system shows +ve deviation as benzene breaks H -bonds among ${CH}_{3} OH$ .

KCET - 2024

CHXII02:SOLUTIONS

319481 The total vapor pressure of a 4 mole% solution of $N H_{3}$ in water at 293 K is 50 torr. The vapor pressure of pure water is 17 torr at this temperature. Applying Henry's law i.e., Solubility $= K_{H} \times$ Pressure and Raoult's law, calculate the total vapour pressure for a 5 mole % solution.

1 33 torr

2 42 torr

3 50.25 torr

4 58.25 torr

Explanation:

According to RLVP, $\frac{P^{0} - P}{P^{0}} = \frac{w_{2} \times M W_{1}}{m w_{2} \times W_{1}}$
4 mole% $N H_{3}$ means 68 g $N H_{3}$ in 1800 g of water.
$\frac{17 - P}{P} = \frac{(68 \times 18)}{(17 \times 1800)}$
P = 0.68 torr
Vapour pressure of $H_{2} O = 17 - 0 .68 = 16 .32 t o r r$
Vapour pressure of $N H_{3} = 50 - 16 .32 = 33 .68 t o r r$
From Henry's law,
$K_{H} = \frac{4}{33 .68} = 0 .118$
5 mole% $N H_{3}$ means 85 g $N H_{3}$ in 1800 g water
$\frac{17 - P}{P} = \frac{(85 \times 18)}{(17 \times 1800)} = 0 .85$
Vapour pressure of water $= 17 - 0 .85 = 16 .15$
From Henry's law,
P = 42.1
Total vapor pressure = $16 .51 + 42 .1 = 58 .25 t o r r$

CHXII02:SOLUTIONS

319482 The solubility of $N_{2}$ in water at 300 K and 500 torr partial pressure is $0 .01 g L^{- 1}$ . The solubility $(i n g L^{- 1})$ at 750 torr partial pressure is :

1 0.02

2 0.015

3 0.0075

4 0.005

Explanation:

According to Henry’s law
$\frac{P_{1}}{P_{2}} = \frac{S_{1}}{S_{2}}$
$∴ S_{1} & S_{2}$ are solubility of gas (g/L)
$\frac{500}{750} = \frac{0 .01}{S_{2}}$
$∴ S_{2} = \frac{750 \times 0 .01}{500} = 0 .015 g / L$ .

JEE - 2016

CHXII02:SOLUTIONS

319483 At $8 0^{^{\circ}} C$ , the vapour pressure of pure liquid ‘A’ is 520 mm Hg and that of pure liquid ‘B’ is 1000 mm Hg. If a mixture of solution of ‘A’ and ‘B’ boils at $8 0^{^{\circ}} C$ and 1 atm pressure, the amount of ‘A’ in the mixture is (1 atm = 760 mm Hg)

1 50 mole percent

2 52 mole percent

3 34 mole percent

4 48 mole percent

Explanation:

$P_{T} {= p}_{A}^{0} X_{A} {+ p}_{B}^{0} X_{B}$
Mixture of solution boils at 1 atm = 760 mm = total pressure
$760 = 520 X_{A} + 1000 (1 - X_{A})$
$X_{A} = 0 .5 m o l % o f A = 50 %$

CHXII02:SOLUTIONS

319480 Which one of the following pairs will show positive deviation from Raoult's Law?

1 Water - HCl

2 Benzene-Methanol

3 Water

- {HNO}_{3}

4 Acetone - Chloroform

Explanation:

Benzene $- {CH}_{3} OH$ system shows +ve deviation as benzene breaks H -bonds among ${CH}_{3} OH$ .

KCET - 2024

CHXII02:SOLUTIONS

319481 The total vapor pressure of a 4 mole% solution of $N H_{3}$ in water at 293 K is 50 torr. The vapor pressure of pure water is 17 torr at this temperature. Applying Henry's law i.e., Solubility $= K_{H} \times$ Pressure and Raoult's law, calculate the total vapour pressure for a 5 mole % solution.

1 33 torr

2 42 torr

3 50.25 torr

4 58.25 torr

Explanation:

According to RLVP, $\frac{P^{0} - P}{P^{0}} = \frac{w_{2} \times M W_{1}}{m w_{2} \times W_{1}}$
4 mole% $N H_{3}$ means 68 g $N H_{3}$ in 1800 g of water.
$\frac{17 - P}{P} = \frac{(68 \times 18)}{(17 \times 1800)}$
P = 0.68 torr
Vapour pressure of $H_{2} O = 17 - 0 .68 = 16 .32 t o r r$
Vapour pressure of $N H_{3} = 50 - 16 .32 = 33 .68 t o r r$
From Henry's law,
$K_{H} = \frac{4}{33 .68} = 0 .118$
5 mole% $N H_{3}$ means 85 g $N H_{3}$ in 1800 g water
$\frac{17 - P}{P} = \frac{(85 \times 18)}{(17 \times 1800)} = 0 .85$
Vapour pressure of water $= 17 - 0 .85 = 16 .15$
From Henry's law,
P = 42.1
Total vapor pressure = $16 .51 + 42 .1 = 58 .25 t o r r$

CHXII02:SOLUTIONS

319482 The solubility of $N_{2}$ in water at 300 K and 500 torr partial pressure is $0 .01 g L^{- 1}$ . The solubility $(i n g L^{- 1})$ at 750 torr partial pressure is :

1 0.02

2 0.015

3 0.0075

4 0.005

Explanation:

According to Henry’s law
$\frac{P_{1}}{P_{2}} = \frac{S_{1}}{S_{2}}$
$∴ S_{1} & S_{2}$ are solubility of gas (g/L)
$\frac{500}{750} = \frac{0 .01}{S_{2}}$
$∴ S_{2} = \frac{750 \times 0 .01}{500} = 0 .015 g / L$ .

JEE - 2016

CHXII02:SOLUTIONS

319483 At $8 0^{^{\circ}} C$ , the vapour pressure of pure liquid ‘A’ is 520 mm Hg and that of pure liquid ‘B’ is 1000 mm Hg. If a mixture of solution of ‘A’ and ‘B’ boils at $8 0^{^{\circ}} C$ and 1 atm pressure, the amount of ‘A’ in the mixture is (1 atm = 760 mm Hg)

1 50 mole percent

2 52 mole percent

3 34 mole percent

4 48 mole percent

Explanation:

$P_{T} {= p}_{A}^{0} X_{A} {+ p}_{B}^{0} X_{B}$
Mixture of solution boils at 1 atm = 760 mm = total pressure
$760 = 520 X_{A} + 1000 (1 - X_{A})$
$X_{A} = 0 .5 m o l % o f A = 50 %$

CHXII02:SOLUTIONS

319480 Which one of the following pairs will show positive deviation from Raoult's Law?

1 Water - HCl

2 Benzene-Methanol

3 Water

- {HNO}_{3}

4 Acetone - Chloroform

Explanation:

Benzene $- {CH}_{3} OH$ system shows +ve deviation as benzene breaks H -bonds among ${CH}_{3} OH$ .

KCET - 2024

CHXII02:SOLUTIONS

319481 The total vapor pressure of a 4 mole% solution of $N H_{3}$ in water at 293 K is 50 torr. The vapor pressure of pure water is 17 torr at this temperature. Applying Henry's law i.e., Solubility $= K_{H} \times$ Pressure and Raoult's law, calculate the total vapour pressure for a 5 mole % solution.

1 33 torr

2 42 torr

3 50.25 torr

4 58.25 torr

Explanation:

According to RLVP, $\frac{P^{0} - P}{P^{0}} = \frac{w_{2} \times M W_{1}}{m w_{2} \times W_{1}}$
4 mole% $N H_{3}$ means 68 g $N H_{3}$ in 1800 g of water.
$\frac{17 - P}{P} = \frac{(68 \times 18)}{(17 \times 1800)}$
P = 0.68 torr
Vapour pressure of $H_{2} O = 17 - 0 .68 = 16 .32 t o r r$
Vapour pressure of $N H_{3} = 50 - 16 .32 = 33 .68 t o r r$
From Henry's law,
$K_{H} = \frac{4}{33 .68} = 0 .118$
5 mole% $N H_{3}$ means 85 g $N H_{3}$ in 1800 g water
$\frac{17 - P}{P} = \frac{(85 \times 18)}{(17 \times 1800)} = 0 .85$
Vapour pressure of water $= 17 - 0 .85 = 16 .15$
From Henry's law,
P = 42.1
Total vapor pressure = $16 .51 + 42 .1 = 58 .25 t o r r$

CHXII02:SOLUTIONS

319482 The solubility of $N_{2}$ in water at 300 K and 500 torr partial pressure is $0 .01 g L^{- 1}$ . The solubility $(i n g L^{- 1})$ at 750 torr partial pressure is :

1 0.02

2 0.015

3 0.0075

4 0.005

Explanation:

According to Henry’s law
$\frac{P_{1}}{P_{2}} = \frac{S_{1}}{S_{2}}$
$∴ S_{1} & S_{2}$ are solubility of gas (g/L)
$\frac{500}{750} = \frac{0 .01}{S_{2}}$
$∴ S_{2} = \frac{750 \times 0 .01}{500} = 0 .015 g / L$ .

JEE - 2016

CHXII02:SOLUTIONS

319483 At $8 0^{^{\circ}} C$ , the vapour pressure of pure liquid ‘A’ is 520 mm Hg and that of pure liquid ‘B’ is 1000 mm Hg. If a mixture of solution of ‘A’ and ‘B’ boils at $8 0^{^{\circ}} C$ and 1 atm pressure, the amount of ‘A’ in the mixture is (1 atm = 760 mm Hg)

1 50 mole percent

2 52 mole percent

3 34 mole percent

4 48 mole percent

Explanation:

$P_{T} {= p}_{A}^{0} X_{A} {+ p}_{B}^{0} X_{B}$
Mixture of solution boils at 1 atm = 760 mm = total pressure
$760 = 520 X_{A} + 1000 (1 - X_{A})$
$X_{A} = 0 .5 m o l % o f A = 50 %$

CHXII02:SOLUTIONS

319480 Which one of the following pairs will show positive deviation from Raoult's Law?

1 Water - HCl

2 Benzene-Methanol

3 Water

- {HNO}_{3}

4 Acetone - Chloroform

Explanation:

Benzene $- {CH}_{3} OH$ system shows +ve deviation as benzene breaks H -bonds among ${CH}_{3} OH$ .

KCET - 2024

CHXII02:SOLUTIONS

319481 The total vapor pressure of a 4 mole% solution of $N H_{3}$ in water at 293 K is 50 torr. The vapor pressure of pure water is 17 torr at this temperature. Applying Henry's law i.e., Solubility $= K_{H} \times$ Pressure and Raoult's law, calculate the total vapour pressure for a 5 mole % solution.

1 33 torr

2 42 torr

3 50.25 torr

4 58.25 torr

Explanation:

According to RLVP, $\frac{P^{0} - P}{P^{0}} = \frac{w_{2} \times M W_{1}}{m w_{2} \times W_{1}}$
4 mole% $N H_{3}$ means 68 g $N H_{3}$ in 1800 g of water.
$\frac{17 - P}{P} = \frac{(68 \times 18)}{(17 \times 1800)}$
P = 0.68 torr
Vapour pressure of $H_{2} O = 17 - 0 .68 = 16 .32 t o r r$
Vapour pressure of $N H_{3} = 50 - 16 .32 = 33 .68 t o r r$
From Henry's law,
$K_{H} = \frac{4}{33 .68} = 0 .118$
5 mole% $N H_{3}$ means 85 g $N H_{3}$ in 1800 g water
$\frac{17 - P}{P} = \frac{(85 \times 18)}{(17 \times 1800)} = 0 .85$
Vapour pressure of water $= 17 - 0 .85 = 16 .15$
From Henry's law,
P = 42.1
Total vapor pressure = $16 .51 + 42 .1 = 58 .25 t o r r$

CHXII02:SOLUTIONS

319482 The solubility of $N_{2}$ in water at 300 K and 500 torr partial pressure is $0 .01 g L^{- 1}$ . The solubility $(i n g L^{- 1})$ at 750 torr partial pressure is :

1 0.02

2 0.015

3 0.0075

4 0.005

Explanation:

According to Henry’s law
$\frac{P_{1}}{P_{2}} = \frac{S_{1}}{S_{2}}$
$∴ S_{1} & S_{2}$ are solubility of gas (g/L)
$\frac{500}{750} = \frac{0 .01}{S_{2}}$
$∴ S_{2} = \frac{750 \times 0 .01}{500} = 0 .015 g / L$ .

JEE - 2016

CHXII02:SOLUTIONS

319483 At $8 0^{^{\circ}} C$ , the vapour pressure of pure liquid ‘A’ is 520 mm Hg and that of pure liquid ‘B’ is 1000 mm Hg. If a mixture of solution of ‘A’ and ‘B’ boils at $8 0^{^{\circ}} C$ and 1 atm pressure, the amount of ‘A’ in the mixture is (1 atm = 760 mm Hg)

1 50 mole percent

2 52 mole percent

3 34 mole percent

4 48 mole percent

Explanation:

$P_{T} {= p}_{A}^{0} X_{A} {+ p}_{B}^{0} X_{B}$
Mixture of solution boils at 1 atm = 760 mm = total pressure
$760 = 520 X_{A} + 1000 (1 - X_{A})$
$X_{A} = 0 .5 m o l % o f A = 50 %$