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CHEMISTRY(KCET)

285361 Relative lowering of vapour pressure of a dilute solution of glucose dissolved in 1 kg of water is 0.002 . The molality of the solution is

1 0.004

2 0.222

3 0.111

4 0.021

Explanation:

(c) Relative lowering of vapour pressure for dilute solution i.e., $\frac{p^{\circ} - p_{s}}{p^{\circ}} = \frac{n_{2}}{n_{1}}$
Given: $\frac{p^{\circ} - p_{s}}{p^{\circ}} = 0.002$ ,
$n_{1 (H_{2} O)} = \frac{1000}{18}$ moles
For dilute solution, $n_{2}$ is molality
$= 0.002 \times \frac{1000}{18} m = 0.111 m$

Karnataka CET 2019

CHEMISTRY(KCET)

285362 A non-volatile solute ' $A$ ' tetramerises in water to the extent of $80 %, 2.5 g$ of ' $A$ ' in 100 g of water, lowers the freezing point by ${0.3}^{\circ} C$ . The molar mass of $A$ in $g {mol}^{- 1}$ is
( $K_{f}$ for water $= 1.86 K kg {mol}^{- 1}$ )

1 62

2 221

3 155

4 354

Explanation:

(a) As non-volatile solute ' $A$ ' tetramerises,
$∴ α (1 - i) \frac{n}{n - 1}, α = 80 %$ (given)
$0.8 = (1 - i) \frac{4}{3} \Rightarrow i = 0.4$
$Δ T_{f} = i K_{f} m$
$0.3 = 0.4 \times 1.86 \times \frac{2.5}{Molar mass} \times \frac{1000}{100}$
Molar mass
$= \frac{0.4 \times 1.86 \times 2.5 \times 10}{0.3} = 62 g / mol$

Karnataka CET 2019

CHEMISTRY(KCET)

285363 Solution ' $A$ ' contains acetone dissolved in chloroform and solution ' $B$ ' contains acetone dissolved in carbon disulphide. The type of deviations from Raoult's law shown by solutions $A$ and $B$ , respectively are

1 positive and positive

2 positive and negative

3 negative and negative

4 negative and positive.

Explanation:

(d) Solution of acetone in chloroform shows negative deviation and in ${CS}_{2}$ it shows positive deviation from Raoult's law.

Karnataka CET 2019

CHEMISTRY(KCET)

285364 Which of the following aqueous solutions should have the highest boiling point?

1 1.0 M NaOH

2

1.0 M {Na}_{2} {SO}_{4}

3

1.0 M {NH}_{4} {NO}_{3}

4

1.0 M {KNO}_{3}

Explanation:

(b) $Δ T_{b} = i K_{b} m$
Concentration is same for all the solutions. So, $Δ T_{b}$ $\propto i$
$i = 2$ for $NaOH; i = 3$ for ${Na}_{2} {SO}_{4}$
$i = 2$ for ${NH}_{4} {NO}_{3}; i = 2$ for ${KNO}_{3}$
So, elevation in boiling point will be maximum in ${Na}_{2} {SO}_{4}$ solution or ${Na}_{2} {SO}_{4}$ solution will have highest boiling point.

Karnataka CET 2018

CHEMISTRY(KCET)

285361 Relative lowering of vapour pressure of a dilute solution of glucose dissolved in 1 kg of water is 0.002 . The molality of the solution is

1 0.004

2 0.222

3 0.111

4 0.021

Explanation:

(c) Relative lowering of vapour pressure for dilute solution i.e., $\frac{p^{\circ} - p_{s}}{p^{\circ}} = \frac{n_{2}}{n_{1}}$
Given: $\frac{p^{\circ} - p_{s}}{p^{\circ}} = 0.002$ ,
$n_{1 (H_{2} O)} = \frac{1000}{18}$ moles
For dilute solution, $n_{2}$ is molality
$= 0.002 \times \frac{1000}{18} m = 0.111 m$

Karnataka CET 2019

CHEMISTRY(KCET)

285362 A non-volatile solute ' $A$ ' tetramerises in water to the extent of $80 %, 2.5 g$ of ' $A$ ' in 100 g of water, lowers the freezing point by ${0.3}^{\circ} C$ . The molar mass of $A$ in $g {mol}^{- 1}$ is
( $K_{f}$ for water $= 1.86 K kg {mol}^{- 1}$ )

1 62

2 221

3 155

4 354

Explanation:

(a) As non-volatile solute ' $A$ ' tetramerises,
$∴ α (1 - i) \frac{n}{n - 1}, α = 80 %$ (given)
$0.8 = (1 - i) \frac{4}{3} \Rightarrow i = 0.4$
$Δ T_{f} = i K_{f} m$
$0.3 = 0.4 \times 1.86 \times \frac{2.5}{Molar mass} \times \frac{1000}{100}$
Molar mass
$= \frac{0.4 \times 1.86 \times 2.5 \times 10}{0.3} = 62 g / mol$

Karnataka CET 2019

CHEMISTRY(KCET)

285363 Solution ' $A$ ' contains acetone dissolved in chloroform and solution ' $B$ ' contains acetone dissolved in carbon disulphide. The type of deviations from Raoult's law shown by solutions $A$ and $B$ , respectively are

1 positive and positive

2 positive and negative

3 negative and negative

4 negative and positive.

Explanation:

(d) Solution of acetone in chloroform shows negative deviation and in ${CS}_{2}$ it shows positive deviation from Raoult's law.

Karnataka CET 2019

CHEMISTRY(KCET)

285364 Which of the following aqueous solutions should have the highest boiling point?

1 1.0 M NaOH

2

1.0 M {Na}_{2} {SO}_{4}

3

1.0 M {NH}_{4} {NO}_{3}

4

1.0 M {KNO}_{3}

Explanation:

(b) $Δ T_{b} = i K_{b} m$
Concentration is same for all the solutions. So, $Δ T_{b}$ $\propto i$
$i = 2$ for $NaOH; i = 3$ for ${Na}_{2} {SO}_{4}$
$i = 2$ for ${NH}_{4} {NO}_{3}; i = 2$ for ${KNO}_{3}$
So, elevation in boiling point will be maximum in ${Na}_{2} {SO}_{4}$ solution or ${Na}_{2} {SO}_{4}$ solution will have highest boiling point.

Karnataka CET 2018

CHEMISTRY(KCET)

285361 Relative lowering of vapour pressure of a dilute solution of glucose dissolved in 1 kg of water is 0.002 . The molality of the solution is

1 0.004

2 0.222

3 0.111

4 0.021

Explanation:

(c) Relative lowering of vapour pressure for dilute solution i.e., $\frac{p^{\circ} - p_{s}}{p^{\circ}} = \frac{n_{2}}{n_{1}}$
Given: $\frac{p^{\circ} - p_{s}}{p^{\circ}} = 0.002$ ,
$n_{1 (H_{2} O)} = \frac{1000}{18}$ moles
For dilute solution, $n_{2}$ is molality
$= 0.002 \times \frac{1000}{18} m = 0.111 m$

Karnataka CET 2019

CHEMISTRY(KCET)

285362 A non-volatile solute ' $A$ ' tetramerises in water to the extent of $80 %, 2.5 g$ of ' $A$ ' in 100 g of water, lowers the freezing point by ${0.3}^{\circ} C$ . The molar mass of $A$ in $g {mol}^{- 1}$ is
( $K_{f}$ for water $= 1.86 K kg {mol}^{- 1}$ )

1 62

2 221

3 155

4 354

Explanation:

(a) As non-volatile solute ' $A$ ' tetramerises,
$∴ α (1 - i) \frac{n}{n - 1}, α = 80 %$ (given)
$0.8 = (1 - i) \frac{4}{3} \Rightarrow i = 0.4$
$Δ T_{f} = i K_{f} m$
$0.3 = 0.4 \times 1.86 \times \frac{2.5}{Molar mass} \times \frac{1000}{100}$
Molar mass
$= \frac{0.4 \times 1.86 \times 2.5 \times 10}{0.3} = 62 g / mol$

Karnataka CET 2019

CHEMISTRY(KCET)

285363 Solution ' $A$ ' contains acetone dissolved in chloroform and solution ' $B$ ' contains acetone dissolved in carbon disulphide. The type of deviations from Raoult's law shown by solutions $A$ and $B$ , respectively are

1 positive and positive

2 positive and negative

3 negative and negative

4 negative and positive.

Explanation:

(d) Solution of acetone in chloroform shows negative deviation and in ${CS}_{2}$ it shows positive deviation from Raoult's law.

Karnataka CET 2019

CHEMISTRY(KCET)

285364 Which of the following aqueous solutions should have the highest boiling point?

1 1.0 M NaOH

2

1.0 M {Na}_{2} {SO}_{4}

3

1.0 M {NH}_{4} {NO}_{3}

4

1.0 M {KNO}_{3}

Explanation:

(b) $Δ T_{b} = i K_{b} m$
Concentration is same for all the solutions. So, $Δ T_{b}$ $\propto i$
$i = 2$ for $NaOH; i = 3$ for ${Na}_{2} {SO}_{4}$
$i = 2$ for ${NH}_{4} {NO}_{3}; i = 2$ for ${KNO}_{3}$
So, elevation in boiling point will be maximum in ${Na}_{2} {SO}_{4}$ solution or ${Na}_{2} {SO}_{4}$ solution will have highest boiling point.

Karnataka CET 2018

CHEMISTRY(KCET)

285361 Relative lowering of vapour pressure of a dilute solution of glucose dissolved in 1 kg of water is 0.002 . The molality of the solution is

1 0.004

2 0.222

3 0.111

4 0.021

Explanation:

(c) Relative lowering of vapour pressure for dilute solution i.e., $\frac{p^{\circ} - p_{s}}{p^{\circ}} = \frac{n_{2}}{n_{1}}$
Given: $\frac{p^{\circ} - p_{s}}{p^{\circ}} = 0.002$ ,
$n_{1 (H_{2} O)} = \frac{1000}{18}$ moles
For dilute solution, $n_{2}$ is molality
$= 0.002 \times \frac{1000}{18} m = 0.111 m$

Karnataka CET 2019

CHEMISTRY(KCET)

285362 A non-volatile solute ' $A$ ' tetramerises in water to the extent of $80 %, 2.5 g$ of ' $A$ ' in 100 g of water, lowers the freezing point by ${0.3}^{\circ} C$ . The molar mass of $A$ in $g {mol}^{- 1}$ is
( $K_{f}$ for water $= 1.86 K kg {mol}^{- 1}$ )

1 62

2 221

3 155

4 354

Explanation:

(a) As non-volatile solute ' $A$ ' tetramerises,
$∴ α (1 - i) \frac{n}{n - 1}, α = 80 %$ (given)
$0.8 = (1 - i) \frac{4}{3} \Rightarrow i = 0.4$
$Δ T_{f} = i K_{f} m$
$0.3 = 0.4 \times 1.86 \times \frac{2.5}{Molar mass} \times \frac{1000}{100}$
Molar mass
$= \frac{0.4 \times 1.86 \times 2.5 \times 10}{0.3} = 62 g / mol$

Karnataka CET 2019

CHEMISTRY(KCET)

285363 Solution ' $A$ ' contains acetone dissolved in chloroform and solution ' $B$ ' contains acetone dissolved in carbon disulphide. The type of deviations from Raoult's law shown by solutions $A$ and $B$ , respectively are

1 positive and positive

2 positive and negative

3 negative and negative

4 negative and positive.

Explanation:

(d) Solution of acetone in chloroform shows negative deviation and in ${CS}_{2}$ it shows positive deviation from Raoult's law.

Karnataka CET 2019

CHEMISTRY(KCET)

285364 Which of the following aqueous solutions should have the highest boiling point?

1 1.0 M NaOH

2

1.0 M {Na}_{2} {SO}_{4}

3

1.0 M {NH}_{4} {NO}_{3}

4

1.0 M {KNO}_{3}

Explanation:

(b) $Δ T_{b} = i K_{b} m$
Concentration is same for all the solutions. So, $Δ T_{b}$ $\propto i$
$i = 2$ for $NaOH; i = 3$ for ${Na}_{2} {SO}_{4}$
$i = 2$ for ${NH}_{4} {NO}_{3}; i = 2$ for ${KNO}_{3}$
So, elevation in boiling point will be maximum in ${Na}_{2} {SO}_{4}$ solution or ${Na}_{2} {SO}_{4}$ solution will have highest boiling point.

Karnataka CET 2018