QBManager

SOLUTIONS

277367 The weight in gram, of the non-volatile solute urea $({NH}_{2} {CONH}_{2})$ to be dissolved in $100 g$ of water in order to decrease its vapour pressure by $1.8 %$ is

1 6.0

2 0.3

3 3.0

4 0.6

Explanation:

According to the Roult's law-
$\frac{P^{o} - P_{s}}{P^{o}} = \frac{n_{2}}{n_{1} + n_{2}}$
Where, $P^{o} =$ Vapour pressure of pure water
$\begin{aligned} n_{2} = Moles of solute \\ n_{1} = Moles of solvent \end{aligned}$
Let vapour pressure of pure water $= 100 mm$ lowering in vapour pressure $= 1.8 mm$ (Given)
$Moles of solute (n_{2}) = \frac{w_{2}}{60} mole$
$Moles of solvent (n_{1}) = \frac{100}{18} mole$
Now, From Rault's law-
$\frac{1.8}{100} = \frac{\frac{w_{2}}{60}}{\frac{w_{2}}{60} + \frac{100}{18}}$
or $\frac{1.8}{100} = \frac{w_{2}}{60 (\frac{3 w_{2} + 1000}{180})}$
or $\frac{1.8}{100} = \frac{w_{2}}{60} \times \frac{180}{(3 w_{2} + 1000)}$
or $\frac{1.8}{100} = \frac{3 w_{2}}{3 w_{2} + 1000}$
or $\frac{100}{1.8} = 1 + \frac{1000}{3 w_{2}}$
or $w_{2} = 6.10$
or $w_{2} \approx 6 g$

AP-EAMCET (Medical)-2013

SOLUTIONS

277368 At a particular temperature, the vapour pressures of two liquids $A$ and $B$ are respectively 120 and $180 mm$ of mercury. If 2 moles of $A$ and 3 moles of $B$ are mixed to form an ideal solution, the vapour pressure of the solution at the same temperature will be (in mm of mercury)

1 156

2 145

3 150

4 108

Explanation:

Given data,
Vapour pressure of liquid $A = 120 mm Hg$
Vapour pressure of liquid $B = 180 mm Hg$
Number of moles of liquid $A = 2$
Number of moles of liquid $B = 3$
Vapour pressure of solution
$\begin{aligned} P_{A} + P_{B} & = X_{A} P_{A}^{o} + X_{B} P_{B}^{o} \\ = \frac{2}{5} \times 120 + \frac{3}{5} \times 180 \\ = 48 + 108 = 156 mm Hg \end{aligned}$

AIIMS-2013

SOLUTIONS

277369 What will be the freezing point of a $1 %$ solution of glucose in water, given that molal depression constant for water is $1.84 K kg$ ${mol}^{- 1}$ ?

1

272.898 K

2

{0.102}^{\circ} C

3

273 K

4

{0.108}^{\circ} C

Explanation:

Mass of glucose $= 1 g$
Mass of solvent $= 100 - 1 = 99 g$
Mass of solution $= 100 g$
So,
$\begin{aligned} = 1.84 \times \frac{1}{180} \times \frac{1}{99 / 100} (∵ Δ T_{f} \times \frac{W}{m} \times \frac{1}{W / 1000}) \\ = 0.103 \end{aligned}$
Now,
Freezing point of solution $= 273 - 0.103$
$= 272.897 K$

J and K CET-(2013)

SOLUTIONS

277371 $0.4 g$ of $H_{2}, 22 g$ of ${CO}_{2}$ and $6.4 g$ of ${SO}_{2}$ are taken in a container. Total pressure is found to be $1.6 atm$ . What is the partial pressure of ${CO}_{2}$ gas in the mixture?

1

1.0 atm

2

0.8 atm

3

0.6 atm

4

1.6 atm

Explanation:

Given that, $P_{Total} = 1.6$ atm
Amount of $H_{2} = 0.4 g$ , moles of $H_{2} = \frac{0.4}{2} = 0.2$ mole
Amount of ${CO}_{2} = 0.22 g$ , moles of ${CO}_{2} = \frac{22}{44} = 0.5 mole$
Amount of ${SO}_{2} = 6.4 g$ , moles of ${SO}_{2} = \frac{6.4}{64} = 0.1 mole$
Now, $P = P_{total} \times X_{{CO}_{2}}$
Where, $P =$ Partial pressure
$P_{Total} =$ Total pressure
$X_{{CO}_{2}} =$ Moles fraction of ${CO}_{2}$
$P_{{CO}_{2}} = 1.6 \times \frac{0.5}{0.5 + 0.2 + 0.1}$
$P_{{CO}_{2}} = 1.6 \times \frac{0.5}{0.8}$
$P_{{CO}_{2}} = 1 atm$

SRMJEEE-2013

SOLUTIONS

277373 $58.4 g$ of $NaCl$ and $180 g$ of glucose were separately dissolved in $1000 ml$ of water. Identify the correct statement regarding the elevation of boiling point (b.p) of the resulting solutions.

1

NaCl

solution will show higher elevation of boiling point

2 Glucose solution will show higher elevation of boiling point

3 Both the solutions will show equal elevation of boiling point

4 The boiling point elevation will be shown by neither of the solutions

Explanation:

Molality of $NaCl = \frac{58.4 g}{58.4 g \times 1000 \times 1} \times 1000 g / kg$
$= 1 m$
For $NaCl$ Solution, using the formula
$\begin{aligned} Δ T_{b} & \propto i \times m \\ = i \times m \\ = 2 \times 1 = 2 \end{aligned}$
Now, Molality of glucose
$\begin{aligned} = \frac{180 g}{180 g / mol \times 1000 ml \times 1 g / ml} \times 1000 \\ = 1 m \end{aligned}$
For glucose $Δ T \propto i \times m$
$= 1 \times 1 = 1$
Now, we can say that $NaCl$ solution will show higher elevation of boiling point.

WB-JEE-2012

SOLUTIONS

277367 The weight in gram, of the non-volatile solute urea $({NH}_{2} {CONH}_{2})$ to be dissolved in $100 g$ of water in order to decrease its vapour pressure by $1.8 %$ is

1 6.0

2 0.3

3 3.0

4 0.6

Explanation:

According to the Roult's law-
$\frac{P^{o} - P_{s}}{P^{o}} = \frac{n_{2}}{n_{1} + n_{2}}$
Where, $P^{o} =$ Vapour pressure of pure water
$\begin{aligned} n_{2} = Moles of solute \\ n_{1} = Moles of solvent \end{aligned}$
Let vapour pressure of pure water $= 100 mm$ lowering in vapour pressure $= 1.8 mm$ (Given)
$Moles of solute (n_{2}) = \frac{w_{2}}{60} mole$
$Moles of solvent (n_{1}) = \frac{100}{18} mole$
Now, From Rault's law-
$\frac{1.8}{100} = \frac{\frac{w_{2}}{60}}{\frac{w_{2}}{60} + \frac{100}{18}}$
or $\frac{1.8}{100} = \frac{w_{2}}{60 (\frac{3 w_{2} + 1000}{180})}$
or $\frac{1.8}{100} = \frac{w_{2}}{60} \times \frac{180}{(3 w_{2} + 1000)}$
or $\frac{1.8}{100} = \frac{3 w_{2}}{3 w_{2} + 1000}$
or $\frac{100}{1.8} = 1 + \frac{1000}{3 w_{2}}$
or $w_{2} = 6.10$
or $w_{2} \approx 6 g$

AP-EAMCET (Medical)-2013

SOLUTIONS

277368 At a particular temperature, the vapour pressures of two liquids $A$ and $B$ are respectively 120 and $180 mm$ of mercury. If 2 moles of $A$ and 3 moles of $B$ are mixed to form an ideal solution, the vapour pressure of the solution at the same temperature will be (in mm of mercury)

1 156

2 145

3 150

4 108

Explanation:

Given data,
Vapour pressure of liquid $A = 120 mm Hg$
Vapour pressure of liquid $B = 180 mm Hg$
Number of moles of liquid $A = 2$
Number of moles of liquid $B = 3$
Vapour pressure of solution
$\begin{aligned} P_{A} + P_{B} & = X_{A} P_{A}^{o} + X_{B} P_{B}^{o} \\ = \frac{2}{5} \times 120 + \frac{3}{5} \times 180 \\ = 48 + 108 = 156 mm Hg \end{aligned}$

AIIMS-2013

SOLUTIONS

277369 What will be the freezing point of a $1 %$ solution of glucose in water, given that molal depression constant for water is $1.84 K kg$ ${mol}^{- 1}$ ?

1

272.898 K

2

{0.102}^{\circ} C

3

273 K

4

{0.108}^{\circ} C

Explanation:

Mass of glucose $= 1 g$
Mass of solvent $= 100 - 1 = 99 g$
Mass of solution $= 100 g$
So,
$\begin{aligned} = 1.84 \times \frac{1}{180} \times \frac{1}{99 / 100} (∵ Δ T_{f} \times \frac{W}{m} \times \frac{1}{W / 1000}) \\ = 0.103 \end{aligned}$
Now,
Freezing point of solution $= 273 - 0.103$
$= 272.897 K$

J and K CET-(2013)

SOLUTIONS

277371 $0.4 g$ of $H_{2}, 22 g$ of ${CO}_{2}$ and $6.4 g$ of ${SO}_{2}$ are taken in a container. Total pressure is found to be $1.6 atm$ . What is the partial pressure of ${CO}_{2}$ gas in the mixture?

1

1.0 atm

2

0.8 atm

3

0.6 atm

4

1.6 atm

Explanation:

Given that, $P_{Total} = 1.6$ atm
Amount of $H_{2} = 0.4 g$ , moles of $H_{2} = \frac{0.4}{2} = 0.2$ mole
Amount of ${CO}_{2} = 0.22 g$ , moles of ${CO}_{2} = \frac{22}{44} = 0.5 mole$
Amount of ${SO}_{2} = 6.4 g$ , moles of ${SO}_{2} = \frac{6.4}{64} = 0.1 mole$
Now, $P = P_{total} \times X_{{CO}_{2}}$
Where, $P =$ Partial pressure
$P_{Total} =$ Total pressure
$X_{{CO}_{2}} =$ Moles fraction of ${CO}_{2}$
$P_{{CO}_{2}} = 1.6 \times \frac{0.5}{0.5 + 0.2 + 0.1}$
$P_{{CO}_{2}} = 1.6 \times \frac{0.5}{0.8}$
$P_{{CO}_{2}} = 1 atm$

SRMJEEE-2013

SOLUTIONS

277373 $58.4 g$ of $NaCl$ and $180 g$ of glucose were separately dissolved in $1000 ml$ of water. Identify the correct statement regarding the elevation of boiling point (b.p) of the resulting solutions.

1

NaCl

solution will show higher elevation of boiling point

2 Glucose solution will show higher elevation of boiling point

3 Both the solutions will show equal elevation of boiling point

4 The boiling point elevation will be shown by neither of the solutions

Explanation:

Molality of $NaCl = \frac{58.4 g}{58.4 g \times 1000 \times 1} \times 1000 g / kg$
$= 1 m$
For $NaCl$ Solution, using the formula
$\begin{aligned} Δ T_{b} & \propto i \times m \\ = i \times m \\ = 2 \times 1 = 2 \end{aligned}$
Now, Molality of glucose
$\begin{aligned} = \frac{180 g}{180 g / mol \times 1000 ml \times 1 g / ml} \times 1000 \\ = 1 m \end{aligned}$
For glucose $Δ T \propto i \times m$
$= 1 \times 1 = 1$
Now, we can say that $NaCl$ solution will show higher elevation of boiling point.

WB-JEE-2012

SOLUTIONS

277367 The weight in gram, of the non-volatile solute urea $({NH}_{2} {CONH}_{2})$ to be dissolved in $100 g$ of water in order to decrease its vapour pressure by $1.8 %$ is

1 6.0

2 0.3

3 3.0

4 0.6

Explanation:

According to the Roult's law-
$\frac{P^{o} - P_{s}}{P^{o}} = \frac{n_{2}}{n_{1} + n_{2}}$
Where, $P^{o} =$ Vapour pressure of pure water
$\begin{aligned} n_{2} = Moles of solute \\ n_{1} = Moles of solvent \end{aligned}$
Let vapour pressure of pure water $= 100 mm$ lowering in vapour pressure $= 1.8 mm$ (Given)
$Moles of solute (n_{2}) = \frac{w_{2}}{60} mole$
$Moles of solvent (n_{1}) = \frac{100}{18} mole$
Now, From Rault's law-
$\frac{1.8}{100} = \frac{\frac{w_{2}}{60}}{\frac{w_{2}}{60} + \frac{100}{18}}$
or $\frac{1.8}{100} = \frac{w_{2}}{60 (\frac{3 w_{2} + 1000}{180})}$
or $\frac{1.8}{100} = \frac{w_{2}}{60} \times \frac{180}{(3 w_{2} + 1000)}$
or $\frac{1.8}{100} = \frac{3 w_{2}}{3 w_{2} + 1000}$
or $\frac{100}{1.8} = 1 + \frac{1000}{3 w_{2}}$
or $w_{2} = 6.10$
or $w_{2} \approx 6 g$

AP-EAMCET (Medical)-2013

SOLUTIONS

277368 At a particular temperature, the vapour pressures of two liquids $A$ and $B$ are respectively 120 and $180 mm$ of mercury. If 2 moles of $A$ and 3 moles of $B$ are mixed to form an ideal solution, the vapour pressure of the solution at the same temperature will be (in mm of mercury)

1 156

2 145

3 150

4 108

Explanation:

Given data,
Vapour pressure of liquid $A = 120 mm Hg$
Vapour pressure of liquid $B = 180 mm Hg$
Number of moles of liquid $A = 2$
Number of moles of liquid $B = 3$
Vapour pressure of solution
$\begin{aligned} P_{A} + P_{B} & = X_{A} P_{A}^{o} + X_{B} P_{B}^{o} \\ = \frac{2}{5} \times 120 + \frac{3}{5} \times 180 \\ = 48 + 108 = 156 mm Hg \end{aligned}$

AIIMS-2013

SOLUTIONS

277369 What will be the freezing point of a $1 %$ solution of glucose in water, given that molal depression constant for water is $1.84 K kg$ ${mol}^{- 1}$ ?

1

272.898 K

2

{0.102}^{\circ} C

3

273 K

4

{0.108}^{\circ} C

Explanation:

Mass of glucose $= 1 g$
Mass of solvent $= 100 - 1 = 99 g$
Mass of solution $= 100 g$
So,
$\begin{aligned} = 1.84 \times \frac{1}{180} \times \frac{1}{99 / 100} (∵ Δ T_{f} \times \frac{W}{m} \times \frac{1}{W / 1000}) \\ = 0.103 \end{aligned}$
Now,
Freezing point of solution $= 273 - 0.103$
$= 272.897 K$

J and K CET-(2013)

SOLUTIONS

277371 $0.4 g$ of $H_{2}, 22 g$ of ${CO}_{2}$ and $6.4 g$ of ${SO}_{2}$ are taken in a container. Total pressure is found to be $1.6 atm$ . What is the partial pressure of ${CO}_{2}$ gas in the mixture?

1

1.0 atm

2

0.8 atm

3

0.6 atm

4

1.6 atm

Explanation:

Given that, $P_{Total} = 1.6$ atm
Amount of $H_{2} = 0.4 g$ , moles of $H_{2} = \frac{0.4}{2} = 0.2$ mole
Amount of ${CO}_{2} = 0.22 g$ , moles of ${CO}_{2} = \frac{22}{44} = 0.5 mole$
Amount of ${SO}_{2} = 6.4 g$ , moles of ${SO}_{2} = \frac{6.4}{64} = 0.1 mole$
Now, $P = P_{total} \times X_{{CO}_{2}}$
Where, $P =$ Partial pressure
$P_{Total} =$ Total pressure
$X_{{CO}_{2}} =$ Moles fraction of ${CO}_{2}$
$P_{{CO}_{2}} = 1.6 \times \frac{0.5}{0.5 + 0.2 + 0.1}$
$P_{{CO}_{2}} = 1.6 \times \frac{0.5}{0.8}$
$P_{{CO}_{2}} = 1 atm$

SRMJEEE-2013

SOLUTIONS

277373 $58.4 g$ of $NaCl$ and $180 g$ of glucose were separately dissolved in $1000 ml$ of water. Identify the correct statement regarding the elevation of boiling point (b.p) of the resulting solutions.

1

NaCl

solution will show higher elevation of boiling point

2 Glucose solution will show higher elevation of boiling point

3 Both the solutions will show equal elevation of boiling point

4 The boiling point elevation will be shown by neither of the solutions

Explanation:

Molality of $NaCl = \frac{58.4 g}{58.4 g \times 1000 \times 1} \times 1000 g / kg$
$= 1 m$
For $NaCl$ Solution, using the formula
$\begin{aligned} Δ T_{b} & \propto i \times m \\ = i \times m \\ = 2 \times 1 = 2 \end{aligned}$
Now, Molality of glucose
$\begin{aligned} = \frac{180 g}{180 g / mol \times 1000 ml \times 1 g / ml} \times 1000 \\ = 1 m \end{aligned}$
For glucose $Δ T \propto i \times m$
$= 1 \times 1 = 1$
Now, we can say that $NaCl$ solution will show higher elevation of boiling point.

WB-JEE-2012

SOLUTIONS

277367 The weight in gram, of the non-volatile solute urea $({NH}_{2} {CONH}_{2})$ to be dissolved in $100 g$ of water in order to decrease its vapour pressure by $1.8 %$ is

1 6.0

2 0.3

3 3.0

4 0.6

Explanation:

According to the Roult's law-
$\frac{P^{o} - P_{s}}{P^{o}} = \frac{n_{2}}{n_{1} + n_{2}}$
Where, $P^{o} =$ Vapour pressure of pure water
$\begin{aligned} n_{2} = Moles of solute \\ n_{1} = Moles of solvent \end{aligned}$
Let vapour pressure of pure water $= 100 mm$ lowering in vapour pressure $= 1.8 mm$ (Given)
$Moles of solute (n_{2}) = \frac{w_{2}}{60} mole$
$Moles of solvent (n_{1}) = \frac{100}{18} mole$
Now, From Rault's law-
$\frac{1.8}{100} = \frac{\frac{w_{2}}{60}}{\frac{w_{2}}{60} + \frac{100}{18}}$
or $\frac{1.8}{100} = \frac{w_{2}}{60 (\frac{3 w_{2} + 1000}{180})}$
or $\frac{1.8}{100} = \frac{w_{2}}{60} \times \frac{180}{(3 w_{2} + 1000)}$
or $\frac{1.8}{100} = \frac{3 w_{2}}{3 w_{2} + 1000}$
or $\frac{100}{1.8} = 1 + \frac{1000}{3 w_{2}}$
or $w_{2} = 6.10$
or $w_{2} \approx 6 g$

AP-EAMCET (Medical)-2013

SOLUTIONS

277368 At a particular temperature, the vapour pressures of two liquids $A$ and $B$ are respectively 120 and $180 mm$ of mercury. If 2 moles of $A$ and 3 moles of $B$ are mixed to form an ideal solution, the vapour pressure of the solution at the same temperature will be (in mm of mercury)

1 156

2 145

3 150

4 108

Explanation:

Given data,
Vapour pressure of liquid $A = 120 mm Hg$
Vapour pressure of liquid $B = 180 mm Hg$
Number of moles of liquid $A = 2$
Number of moles of liquid $B = 3$
Vapour pressure of solution
$\begin{aligned} P_{A} + P_{B} & = X_{A} P_{A}^{o} + X_{B} P_{B}^{o} \\ = \frac{2}{5} \times 120 + \frac{3}{5} \times 180 \\ = 48 + 108 = 156 mm Hg \end{aligned}$

AIIMS-2013

SOLUTIONS

277369 What will be the freezing point of a $1 %$ solution of glucose in water, given that molal depression constant for water is $1.84 K kg$ ${mol}^{- 1}$ ?

1

272.898 K

2

{0.102}^{\circ} C

3

273 K

4

{0.108}^{\circ} C

Explanation:

Mass of glucose $= 1 g$
Mass of solvent $= 100 - 1 = 99 g$
Mass of solution $= 100 g$
So,
$\begin{aligned} = 1.84 \times \frac{1}{180} \times \frac{1}{99 / 100} (∵ Δ T_{f} \times \frac{W}{m} \times \frac{1}{W / 1000}) \\ = 0.103 \end{aligned}$
Now,
Freezing point of solution $= 273 - 0.103$
$= 272.897 K$

J and K CET-(2013)

SOLUTIONS

277371 $0.4 g$ of $H_{2}, 22 g$ of ${CO}_{2}$ and $6.4 g$ of ${SO}_{2}$ are taken in a container. Total pressure is found to be $1.6 atm$ . What is the partial pressure of ${CO}_{2}$ gas in the mixture?

1

1.0 atm

2

0.8 atm

3

0.6 atm

4

1.6 atm

Explanation:

Given that, $P_{Total} = 1.6$ atm
Amount of $H_{2} = 0.4 g$ , moles of $H_{2} = \frac{0.4}{2} = 0.2$ mole
Amount of ${CO}_{2} = 0.22 g$ , moles of ${CO}_{2} = \frac{22}{44} = 0.5 mole$
Amount of ${SO}_{2} = 6.4 g$ , moles of ${SO}_{2} = \frac{6.4}{64} = 0.1 mole$
Now, $P = P_{total} \times X_{{CO}_{2}}$
Where, $P =$ Partial pressure
$P_{Total} =$ Total pressure
$X_{{CO}_{2}} =$ Moles fraction of ${CO}_{2}$
$P_{{CO}_{2}} = 1.6 \times \frac{0.5}{0.5 + 0.2 + 0.1}$
$P_{{CO}_{2}} = 1.6 \times \frac{0.5}{0.8}$
$P_{{CO}_{2}} = 1 atm$

SRMJEEE-2013

SOLUTIONS

277373 $58.4 g$ of $NaCl$ and $180 g$ of glucose were separately dissolved in $1000 ml$ of water. Identify the correct statement regarding the elevation of boiling point (b.p) of the resulting solutions.

1

NaCl

solution will show higher elevation of boiling point

2 Glucose solution will show higher elevation of boiling point

3 Both the solutions will show equal elevation of boiling point

4 The boiling point elevation will be shown by neither of the solutions

Explanation:

Molality of $NaCl = \frac{58.4 g}{58.4 g \times 1000 \times 1} \times 1000 g / kg$
$= 1 m$
For $NaCl$ Solution, using the formula
$\begin{aligned} Δ T_{b} & \propto i \times m \\ = i \times m \\ = 2 \times 1 = 2 \end{aligned}$
Now, Molality of glucose
$\begin{aligned} = \frac{180 g}{180 g / mol \times 1000 ml \times 1 g / ml} \times 1000 \\ = 1 m \end{aligned}$
For glucose $Δ T \propto i \times m$
$= 1 \times 1 = 1$
Now, we can say that $NaCl$ solution will show higher elevation of boiling point.

WB-JEE-2012

SOLUTIONS

277367 The weight in gram, of the non-volatile solute urea $({NH}_{2} {CONH}_{2})$ to be dissolved in $100 g$ of water in order to decrease its vapour pressure by $1.8 %$ is

1 6.0

2 0.3

3 3.0

4 0.6

Explanation:

According to the Roult's law-
$\frac{P^{o} - P_{s}}{P^{o}} = \frac{n_{2}}{n_{1} + n_{2}}$
Where, $P^{o} =$ Vapour pressure of pure water
$\begin{aligned} n_{2} = Moles of solute \\ n_{1} = Moles of solvent \end{aligned}$
Let vapour pressure of pure water $= 100 mm$ lowering in vapour pressure $= 1.8 mm$ (Given)
$Moles of solute (n_{2}) = \frac{w_{2}}{60} mole$
$Moles of solvent (n_{1}) = \frac{100}{18} mole$
Now, From Rault's law-
$\frac{1.8}{100} = \frac{\frac{w_{2}}{60}}{\frac{w_{2}}{60} + \frac{100}{18}}$
or $\frac{1.8}{100} = \frac{w_{2}}{60 (\frac{3 w_{2} + 1000}{180})}$
or $\frac{1.8}{100} = \frac{w_{2}}{60} \times \frac{180}{(3 w_{2} + 1000)}$
or $\frac{1.8}{100} = \frac{3 w_{2}}{3 w_{2} + 1000}$
or $\frac{100}{1.8} = 1 + \frac{1000}{3 w_{2}}$
or $w_{2} = 6.10$
or $w_{2} \approx 6 g$

AP-EAMCET (Medical)-2013

SOLUTIONS

277368 At a particular temperature, the vapour pressures of two liquids $A$ and $B$ are respectively 120 and $180 mm$ of mercury. If 2 moles of $A$ and 3 moles of $B$ are mixed to form an ideal solution, the vapour pressure of the solution at the same temperature will be (in mm of mercury)

1 156

2 145

3 150

4 108

Explanation:

Given data,
Vapour pressure of liquid $A = 120 mm Hg$
Vapour pressure of liquid $B = 180 mm Hg$
Number of moles of liquid $A = 2$
Number of moles of liquid $B = 3$
Vapour pressure of solution
$\begin{aligned} P_{A} + P_{B} & = X_{A} P_{A}^{o} + X_{B} P_{B}^{o} \\ = \frac{2}{5} \times 120 + \frac{3}{5} \times 180 \\ = 48 + 108 = 156 mm Hg \end{aligned}$

AIIMS-2013

SOLUTIONS

277369 What will be the freezing point of a $1 %$ solution of glucose in water, given that molal depression constant for water is $1.84 K kg$ ${mol}^{- 1}$ ?

1

272.898 K

2

{0.102}^{\circ} C

3

273 K

4

{0.108}^{\circ} C

Explanation:

Mass of glucose $= 1 g$
Mass of solvent $= 100 - 1 = 99 g$
Mass of solution $= 100 g$
So,
$\begin{aligned} = 1.84 \times \frac{1}{180} \times \frac{1}{99 / 100} (∵ Δ T_{f} \times \frac{W}{m} \times \frac{1}{W / 1000}) \\ = 0.103 \end{aligned}$
Now,
Freezing point of solution $= 273 - 0.103$
$= 272.897 K$

J and K CET-(2013)

SOLUTIONS

277371 $0.4 g$ of $H_{2}, 22 g$ of ${CO}_{2}$ and $6.4 g$ of ${SO}_{2}$ are taken in a container. Total pressure is found to be $1.6 atm$ . What is the partial pressure of ${CO}_{2}$ gas in the mixture?

1

1.0 atm

2

0.8 atm

3

0.6 atm

4

1.6 atm

Explanation:

Given that, $P_{Total} = 1.6$ atm
Amount of $H_{2} = 0.4 g$ , moles of $H_{2} = \frac{0.4}{2} = 0.2$ mole
Amount of ${CO}_{2} = 0.22 g$ , moles of ${CO}_{2} = \frac{22}{44} = 0.5 mole$
Amount of ${SO}_{2} = 6.4 g$ , moles of ${SO}_{2} = \frac{6.4}{64} = 0.1 mole$
Now, $P = P_{total} \times X_{{CO}_{2}}$
Where, $P =$ Partial pressure
$P_{Total} =$ Total pressure
$X_{{CO}_{2}} =$ Moles fraction of ${CO}_{2}$
$P_{{CO}_{2}} = 1.6 \times \frac{0.5}{0.5 + 0.2 + 0.1}$
$P_{{CO}_{2}} = 1.6 \times \frac{0.5}{0.8}$
$P_{{CO}_{2}} = 1 atm$

SRMJEEE-2013

SOLUTIONS

277373 $58.4 g$ of $NaCl$ and $180 g$ of glucose were separately dissolved in $1000 ml$ of water. Identify the correct statement regarding the elevation of boiling point (b.p) of the resulting solutions.

1

NaCl

solution will show higher elevation of boiling point

2 Glucose solution will show higher elevation of boiling point

3 Both the solutions will show equal elevation of boiling point

4 The boiling point elevation will be shown by neither of the solutions

Explanation:

Molality of $NaCl = \frac{58.4 g}{58.4 g \times 1000 \times 1} \times 1000 g / kg$
$= 1 m$
For $NaCl$ Solution, using the formula
$\begin{aligned} Δ T_{b} & \propto i \times m \\ = i \times m \\ = 2 \times 1 = 2 \end{aligned}$
Now, Molality of glucose
$\begin{aligned} = \frac{180 g}{180 g / mol \times 1000 ml \times 1 g / ml} \times 1000 \\ = 1 m \end{aligned}$
For glucose $Δ T \propto i \times m$
$= 1 \times 1 = 1$
Now, we can say that $NaCl$ solution will show higher elevation of boiling point.

WB-JEE-2012