QBManager

SOLUTIONS

277474 If $20 g$ of a solute was dissolve in $500 mL$ of water and osmotic pressure of the solution was found to be $600 mm$ of $Hg$ at $15^{\circ} C$ , then molecular weight of the solute is

1 1000

2 1200

3 1400

4 1800

Explanation:

Mass of solution $(m) = 20 g$
Volume of the solution $(V) = 500 mL = 0.5 L$
Osmotic pressure of solute $(P) = 600 mm$ of $Hg$
$P = \frac{600}{760} = 0.7895 atm$
From formula,
$Π V = nRT$
$Π = \frac{n}{V} RT = \frac{mRT}{MV}$
Putting the value
$\frac{600}{760} atm = \frac{20 \times 0.0821 \times (273 + 15)}{M \times 0.5 L}$
So,
$M = 1200$

UPTU/UPSEE-2012

SOLUTIONS

277475 $0.1 M NaCl$ and $0.1 M {CH}_{3} COOH$ are kept in separate containers. If their osmotic pressures are $p_{1}$ and $p_{2}$ respectively then what is the correct statement?

1

p_{1} > p_{2}

2

p_{1} = p_{2}

3

p_{1} < p_{2}

4

p_{1} = p_{2} = 0 atm

Explanation:

$\begin{aligned} ended with \end{array}$

JCECE - 2011

SOLUTIONS

277476 Assuming the salts to unionised solution which of the following has highest osmotic pressure?

1

1 % CsCl

2

1 % RbCl

3

1 % KCl

4

1 % NaCl

Explanation:

$∵$ Salts are unionised
$∴ i$ is same for all
Now, $Π = CRT$
$\begin{aligned} Π = \frac{w}{m \times v} RT \\ Π \propto \frac{1}{m} \end{aligned}$
$1 %$ of salt means 1 gram solute dissolved in 100 gram of solvent.
$∵ NaCl$ have lowest molecules mass.
Hence, has highest osmotic pressure.

AMU-2011

SOLUTIONS

277479 Vapour pressure of pure ' $A$ ' is $70 mm of Hg$ at $20^{\circ} C$ . It forms an ideal solution with ' $B$ ' in which mole fraction of $A$ is 0.8 . If the vapour pressure of the solution is $84 mm$ of $Hg$ at $25^{\circ} C$ , the vapour pressure of pure ' $B$ ' at $25^{\circ} C$ is

1

28 mm

2

56 mm

3

70 mm

4

140 mm

Explanation:

Vapour pressure of pure 'A', $P_{A}^{\circ} = 70 mm$ of $Hg$
Mole fraction of $A = 0.8$
Mole fraction of $B = 0.2$
Then,
Vapour pressure of solution, $(P) = 84 mm$
$p = p_{A}^{o} x_{A} + p_{b}^{o} x_{B}$
$\Rightarrow 84 = 70 \times 0.8 + p_{B}^{o} \times 0.2$
$84 = 56 + p_{B}^{o} \times 0.2$
$p_{B}^{o} = \frac{28}{0.2} = 140 mm$

KARNATAKA-CET

SOLUTIONS

277474 If $20 g$ of a solute was dissolve in $500 mL$ of water and osmotic pressure of the solution was found to be $600 mm$ of $Hg$ at $15^{\circ} C$ , then molecular weight of the solute is

1 1000

2 1200

3 1400

4 1800

Explanation:

Mass of solution $(m) = 20 g$
Volume of the solution $(V) = 500 mL = 0.5 L$
Osmotic pressure of solute $(P) = 600 mm$ of $Hg$
$P = \frac{600}{760} = 0.7895 atm$
From formula,
$Π V = nRT$
$Π = \frac{n}{V} RT = \frac{mRT}{MV}$
Putting the value
$\frac{600}{760} atm = \frac{20 \times 0.0821 \times (273 + 15)}{M \times 0.5 L}$
So,
$M = 1200$

UPTU/UPSEE-2012

SOLUTIONS

277475 $0.1 M NaCl$ and $0.1 M {CH}_{3} COOH$ are kept in separate containers. If their osmotic pressures are $p_{1}$ and $p_{2}$ respectively then what is the correct statement?

1

p_{1} > p_{2}

2

p_{1} = p_{2}

3

p_{1} < p_{2}

4

p_{1} = p_{2} = 0 atm

Explanation:

$\begin{aligned} ended with \end{array}$

JCECE - 2011

SOLUTIONS

277476 Assuming the salts to unionised solution which of the following has highest osmotic pressure?

1

1 % CsCl

2

1 % RbCl

3

1 % KCl

4

1 % NaCl

Explanation:

$∵$ Salts are unionised
$∴ i$ is same for all
Now, $Π = CRT$
$\begin{aligned} Π = \frac{w}{m \times v} RT \\ Π \propto \frac{1}{m} \end{aligned}$
$1 %$ of salt means 1 gram solute dissolved in 100 gram of solvent.
$∵ NaCl$ have lowest molecules mass.
Hence, has highest osmotic pressure.

AMU-2011

SOLUTIONS

277479 Vapour pressure of pure ' $A$ ' is $70 mm of Hg$ at $20^{\circ} C$ . It forms an ideal solution with ' $B$ ' in which mole fraction of $A$ is 0.8 . If the vapour pressure of the solution is $84 mm$ of $Hg$ at $25^{\circ} C$ , the vapour pressure of pure ' $B$ ' at $25^{\circ} C$ is

1

28 mm

2

56 mm

3

70 mm

4

140 mm

Explanation:

Vapour pressure of pure 'A', $P_{A}^{\circ} = 70 mm$ of $Hg$
Mole fraction of $A = 0.8$
Mole fraction of $B = 0.2$
Then,
Vapour pressure of solution, $(P) = 84 mm$
$p = p_{A}^{o} x_{A} + p_{b}^{o} x_{B}$
$\Rightarrow 84 = 70 \times 0.8 + p_{B}^{o} \times 0.2$
$84 = 56 + p_{B}^{o} \times 0.2$
$p_{B}^{o} = \frac{28}{0.2} = 140 mm$

KARNATAKA-CET

SOLUTIONS

277474 If $20 g$ of a solute was dissolve in $500 mL$ of water and osmotic pressure of the solution was found to be $600 mm$ of $Hg$ at $15^{\circ} C$ , then molecular weight of the solute is

1 1000

2 1200

3 1400

4 1800

Explanation:

Mass of solution $(m) = 20 g$
Volume of the solution $(V) = 500 mL = 0.5 L$
Osmotic pressure of solute $(P) = 600 mm$ of $Hg$
$P = \frac{600}{760} = 0.7895 atm$
From formula,
$Π V = nRT$
$Π = \frac{n}{V} RT = \frac{mRT}{MV}$
Putting the value
$\frac{600}{760} atm = \frac{20 \times 0.0821 \times (273 + 15)}{M \times 0.5 L}$
So,
$M = 1200$

UPTU/UPSEE-2012

SOLUTIONS

277475 $0.1 M NaCl$ and $0.1 M {CH}_{3} COOH$ are kept in separate containers. If their osmotic pressures are $p_{1}$ and $p_{2}$ respectively then what is the correct statement?

1

p_{1} > p_{2}

2

p_{1} = p_{2}

3

p_{1} < p_{2}

4

p_{1} = p_{2} = 0 atm

Explanation:

$\begin{aligned} ended with \end{array}$

JCECE - 2011

SOLUTIONS

277476 Assuming the salts to unionised solution which of the following has highest osmotic pressure?

1

1 % CsCl

2

1 % RbCl

3

1 % KCl

4

1 % NaCl

Explanation:

$∵$ Salts are unionised
$∴ i$ is same for all
Now, $Π = CRT$
$\begin{aligned} Π = \frac{w}{m \times v} RT \\ Π \propto \frac{1}{m} \end{aligned}$
$1 %$ of salt means 1 gram solute dissolved in 100 gram of solvent.
$∵ NaCl$ have lowest molecules mass.
Hence, has highest osmotic pressure.

AMU-2011

SOLUTIONS

277479 Vapour pressure of pure ' $A$ ' is $70 mm of Hg$ at $20^{\circ} C$ . It forms an ideal solution with ' $B$ ' in which mole fraction of $A$ is 0.8 . If the vapour pressure of the solution is $84 mm$ of $Hg$ at $25^{\circ} C$ , the vapour pressure of pure ' $B$ ' at $25^{\circ} C$ is

1

28 mm

2

56 mm

3

70 mm

4

140 mm

Explanation:

Vapour pressure of pure 'A', $P_{A}^{\circ} = 70 mm$ of $Hg$
Mole fraction of $A = 0.8$
Mole fraction of $B = 0.2$
Then,
Vapour pressure of solution, $(P) = 84 mm$
$p = p_{A}^{o} x_{A} + p_{b}^{o} x_{B}$
$\Rightarrow 84 = 70 \times 0.8 + p_{B}^{o} \times 0.2$
$84 = 56 + p_{B}^{o} \times 0.2$
$p_{B}^{o} = \frac{28}{0.2} = 140 mm$

KARNATAKA-CET

SOLUTIONS

277474 If $20 g$ of a solute was dissolve in $500 mL$ of water and osmotic pressure of the solution was found to be $600 mm$ of $Hg$ at $15^{\circ} C$ , then molecular weight of the solute is

1 1000

2 1200

3 1400

4 1800

Explanation:

Mass of solution $(m) = 20 g$
Volume of the solution $(V) = 500 mL = 0.5 L$
Osmotic pressure of solute $(P) = 600 mm$ of $Hg$
$P = \frac{600}{760} = 0.7895 atm$
From formula,
$Π V = nRT$
$Π = \frac{n}{V} RT = \frac{mRT}{MV}$
Putting the value
$\frac{600}{760} atm = \frac{20 \times 0.0821 \times (273 + 15)}{M \times 0.5 L}$
So,
$M = 1200$

UPTU/UPSEE-2012

SOLUTIONS

277475 $0.1 M NaCl$ and $0.1 M {CH}_{3} COOH$ are kept in separate containers. If their osmotic pressures are $p_{1}$ and $p_{2}$ respectively then what is the correct statement?

1

p_{1} > p_{2}

2

p_{1} = p_{2}

3

p_{1} < p_{2}

4

p_{1} = p_{2} = 0 atm

Explanation:

$\begin{aligned} ended with \end{array}$

JCECE - 2011

SOLUTIONS

277476 Assuming the salts to unionised solution which of the following has highest osmotic pressure?

1

1 % CsCl

2

1 % RbCl

3

1 % KCl

4

1 % NaCl

Explanation:

$∵$ Salts are unionised
$∴ i$ is same for all
Now, $Π = CRT$
$\begin{aligned} Π = \frac{w}{m \times v} RT \\ Π \propto \frac{1}{m} \end{aligned}$
$1 %$ of salt means 1 gram solute dissolved in 100 gram of solvent.
$∵ NaCl$ have lowest molecules mass.
Hence, has highest osmotic pressure.

AMU-2011

SOLUTIONS

277479 Vapour pressure of pure ' $A$ ' is $70 mm of Hg$ at $20^{\circ} C$ . It forms an ideal solution with ' $B$ ' in which mole fraction of $A$ is 0.8 . If the vapour pressure of the solution is $84 mm$ of $Hg$ at $25^{\circ} C$ , the vapour pressure of pure ' $B$ ' at $25^{\circ} C$ is

1

28 mm

2

56 mm

3

70 mm

4

140 mm

Explanation:

Vapour pressure of pure 'A', $P_{A}^{\circ} = 70 mm$ of $Hg$
Mole fraction of $A = 0.8$
Mole fraction of $B = 0.2$
Then,
Vapour pressure of solution, $(P) = 84 mm$
$p = p_{A}^{o} x_{A} + p_{b}^{o} x_{B}$
$\Rightarrow 84 = 70 \times 0.8 + p_{B}^{o} \times 0.2$
$84 = 56 + p_{B}^{o} \times 0.2$
$p_{B}^{o} = \frac{28}{0.2} = 140 mm$

KARNATAKA-CET