QBManager

CHEMISTRY(KCET)

285367 When the pure solvent diffuses out of the solution through the semipermeable membrane then the process is called

1 sorption

2 dialysis

3 osmosis

4 reverse osmosis.

Karnataka CET 2017

CHEMISTRY(KCET)

285368 The van't Hoff factor ' $i$ ' accounts for

1 extent of dissolution of solute

2 extent of dissociation of solute

3 extent of mobility of solute

4 extent of solubility of solute.

Karnataka CET 2017

CHEMISTRY(KCET)

285359 Which of the following pair of solutions is isotonic?

1

0.001 M {Al}_{2} {({SO}_{4})}_{3}

and

0.01 M {BaCl}_{2}

2

0.001 M {CaCl}_{2}

and

0.001 M {Al}_{2} {({SO}_{4})}_{3}

3

0.01 M {BaCl}_{2}

and

0.001 M {CaCl}_{2}

4

0.01 M {BaCl}_{2}

and 0.015 M NaCl

Explanation:

(d) Isotonic solutions are those which have same osmotic pressure at a given temperature.
For $0.01 M {BaCl}_{2}, i = 3$
$π = i C R T = 3 \times 0.01 R T = 0.03 R T$
For $0.015 M NaCl, i = 2$
$∴ π = 2 \times 0.015 R T = 0.03 R T$
Therefore, $0.01 M {BaCl}_{2}$ and 0.015 M NaCl are isotonic.

Karnataka CET 2020

CHEMISTRY(KCET)

285360 Solute ' $X$ ' dimerises in water to the extent of $80 %$ , 2.5 g of $^{'} X$ ' in 100 g of water increases the boiling point by ${0.3}^{\circ} C$ . The molar mass of ' $X$ ' is $[K_{b} = 0.52$ $K kg {mol}^{- 1}$ ]

1 52

2 65

3 26

4 13

Explanation:

(c) Given : $W_{B} = 2.5 g$ ;
$W_{A} = 100 g, K_{b} = 0.52 K kg {mol}^{- 1}$ ,
$Δ T_{b} = {0.3}^{\circ} C$
van't Hoff factor,
$i = 1 + α (\frac{1}{n} - 1)$
$i = 1 + 0.8 (\frac{1}{2} - 1)$
$i = 1 - 0.4 = 0.6$
As $Δ T_{b} = i K_{b} m$
$= i \times K_{b} \times \frac{W_{B} \times 1000}{M_{B} \times W_{A}}$
$0.3 = 0.6 \times 0.52 \times \frac{2.5}{M_{B}} \times \frac{1000}{100}$
$M_{B} = \frac{0.6 \times 0.52 \times 2.5 \times 1000}{0.3 \times 100} = 26 g / mol$

Karnataka CET 2020

CHEMISTRY(KCET)

285367 When the pure solvent diffuses out of the solution through the semipermeable membrane then the process is called

1 sorption

2 dialysis

3 osmosis

4 reverse osmosis.

Karnataka CET 2017

CHEMISTRY(KCET)

285368 The van't Hoff factor ' $i$ ' accounts for

1 extent of dissolution of solute

2 extent of dissociation of solute

3 extent of mobility of solute

4 extent of solubility of solute.

Karnataka CET 2017

CHEMISTRY(KCET)

285359 Which of the following pair of solutions is isotonic?

1

0.001 M {Al}_{2} {({SO}_{4})}_{3}

and

0.01 M {BaCl}_{2}

2

0.001 M {CaCl}_{2}

and

0.001 M {Al}_{2} {({SO}_{4})}_{3}

3

0.01 M {BaCl}_{2}

and

0.001 M {CaCl}_{2}

4

0.01 M {BaCl}_{2}

and 0.015 M NaCl

Explanation:

(d) Isotonic solutions are those which have same osmotic pressure at a given temperature.
For $0.01 M {BaCl}_{2}, i = 3$
$π = i C R T = 3 \times 0.01 R T = 0.03 R T$
For $0.015 M NaCl, i = 2$
$∴ π = 2 \times 0.015 R T = 0.03 R T$
Therefore, $0.01 M {BaCl}_{2}$ and 0.015 M NaCl are isotonic.

Karnataka CET 2020

CHEMISTRY(KCET)

285360 Solute ' $X$ ' dimerises in water to the extent of $80 %$ , 2.5 g of $^{'} X$ ' in 100 g of water increases the boiling point by ${0.3}^{\circ} C$ . The molar mass of ' $X$ ' is $[K_{b} = 0.52$ $K kg {mol}^{- 1}$ ]

1 52

2 65

3 26

4 13

Explanation:

(c) Given : $W_{B} = 2.5 g$ ;
$W_{A} = 100 g, K_{b} = 0.52 K kg {mol}^{- 1}$ ,
$Δ T_{b} = {0.3}^{\circ} C$
van't Hoff factor,
$i = 1 + α (\frac{1}{n} - 1)$
$i = 1 + 0.8 (\frac{1}{2} - 1)$
$i = 1 - 0.4 = 0.6$
As $Δ T_{b} = i K_{b} m$
$= i \times K_{b} \times \frac{W_{B} \times 1000}{M_{B} \times W_{A}}$
$0.3 = 0.6 \times 0.52 \times \frac{2.5}{M_{B}} \times \frac{1000}{100}$
$M_{B} = \frac{0.6 \times 0.52 \times 2.5 \times 1000}{0.3 \times 100} = 26 g / mol$

Karnataka CET 2020

CHEMISTRY(KCET)

285367 When the pure solvent diffuses out of the solution through the semipermeable membrane then the process is called

1 sorption

2 dialysis

3 osmosis

4 reverse osmosis.

Karnataka CET 2017

CHEMISTRY(KCET)

285368 The van't Hoff factor ' $i$ ' accounts for

1 extent of dissolution of solute

2 extent of dissociation of solute

3 extent of mobility of solute

4 extent of solubility of solute.

Karnataka CET 2017

CHEMISTRY(KCET)

285359 Which of the following pair of solutions is isotonic?

1

0.001 M {Al}_{2} {({SO}_{4})}_{3}

and

0.01 M {BaCl}_{2}

2

0.001 M {CaCl}_{2}

and

0.001 M {Al}_{2} {({SO}_{4})}_{3}

3

0.01 M {BaCl}_{2}

and

0.001 M {CaCl}_{2}

4

0.01 M {BaCl}_{2}

and 0.015 M NaCl

Explanation:

(d) Isotonic solutions are those which have same osmotic pressure at a given temperature.
For $0.01 M {BaCl}_{2}, i = 3$
$π = i C R T = 3 \times 0.01 R T = 0.03 R T$
For $0.015 M NaCl, i = 2$
$∴ π = 2 \times 0.015 R T = 0.03 R T$
Therefore, $0.01 M {BaCl}_{2}$ and 0.015 M NaCl are isotonic.

Karnataka CET 2020

CHEMISTRY(KCET)

285360 Solute ' $X$ ' dimerises in water to the extent of $80 %$ , 2.5 g of $^{'} X$ ' in 100 g of water increases the boiling point by ${0.3}^{\circ} C$ . The molar mass of ' $X$ ' is $[K_{b} = 0.52$ $K kg {mol}^{- 1}$ ]

1 52

2 65

3 26

4 13

Explanation:

(c) Given : $W_{B} = 2.5 g$ ;
$W_{A} = 100 g, K_{b} = 0.52 K kg {mol}^{- 1}$ ,
$Δ T_{b} = {0.3}^{\circ} C$
van't Hoff factor,
$i = 1 + α (\frac{1}{n} - 1)$
$i = 1 + 0.8 (\frac{1}{2} - 1)$
$i = 1 - 0.4 = 0.6$
As $Δ T_{b} = i K_{b} m$
$= i \times K_{b} \times \frac{W_{B} \times 1000}{M_{B} \times W_{A}}$
$0.3 = 0.6 \times 0.52 \times \frac{2.5}{M_{B}} \times \frac{1000}{100}$
$M_{B} = \frac{0.6 \times 0.52 \times 2.5 \times 1000}{0.3 \times 100} = 26 g / mol$

Karnataka CET 2020

CHEMISTRY(KCET)

285367 When the pure solvent diffuses out of the solution through the semipermeable membrane then the process is called

1 sorption

2 dialysis

3 osmosis

4 reverse osmosis.

Karnataka CET 2017

CHEMISTRY(KCET)

285368 The van't Hoff factor ' $i$ ' accounts for

1 extent of dissolution of solute

2 extent of dissociation of solute

3 extent of mobility of solute

4 extent of solubility of solute.

Karnataka CET 2017

CHEMISTRY(KCET)

285359 Which of the following pair of solutions is isotonic?

1

0.001 M {Al}_{2} {({SO}_{4})}_{3}

and

0.01 M {BaCl}_{2}

2

0.001 M {CaCl}_{2}

and

0.001 M {Al}_{2} {({SO}_{4})}_{3}

3

0.01 M {BaCl}_{2}

and

0.001 M {CaCl}_{2}

4

0.01 M {BaCl}_{2}

and 0.015 M NaCl

Explanation:

(d) Isotonic solutions are those which have same osmotic pressure at a given temperature.
For $0.01 M {BaCl}_{2}, i = 3$
$π = i C R T = 3 \times 0.01 R T = 0.03 R T$
For $0.015 M NaCl, i = 2$
$∴ π = 2 \times 0.015 R T = 0.03 R T$
Therefore, $0.01 M {BaCl}_{2}$ and 0.015 M NaCl are isotonic.

Karnataka CET 2020

CHEMISTRY(KCET)

285360 Solute ' $X$ ' dimerises in water to the extent of $80 %$ , 2.5 g of $^{'} X$ ' in 100 g of water increases the boiling point by ${0.3}^{\circ} C$ . The molar mass of ' $X$ ' is $[K_{b} = 0.52$ $K kg {mol}^{- 1}$ ]

1 52

2 65

3 26

4 13

Explanation:

(c) Given : $W_{B} = 2.5 g$ ;
$W_{A} = 100 g, K_{b} = 0.52 K kg {mol}^{- 1}$ ,
$Δ T_{b} = {0.3}^{\circ} C$
van't Hoff factor,
$i = 1 + α (\frac{1}{n} - 1)$
$i = 1 + 0.8 (\frac{1}{2} - 1)$
$i = 1 - 0.4 = 0.6$
As $Δ T_{b} = i K_{b} m$
$= i \times K_{b} \times \frac{W_{B} \times 1000}{M_{B} \times W_{A}}$
$0.3 = 0.6 \times 0.52 \times \frac{2.5}{M_{B}} \times \frac{1000}{100}$
$M_{B} = \frac{0.6 \times 0.52 \times 2.5 \times 1000}{0.3 \times 100} = 26 g / mol$

Karnataka CET 2020

Question Bank Series

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