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CHXII02:SOLUTIONS

319088 The freezing point depression of a $1.00 \times 10^{- 3}$ molal aqueous aqueous solution of $K_{x} [F e {(C N)}_{6}]$ is $7.10 \times 10^{- 3} K .$ If $K_{f} (H_{2} O)$ $= 1.86 K K g m o l^{- 1},$ what is the value of x ?

1 5

2 8

3 3

4 4

Explanation:

$K_{x} [Fe {(CN)}_{6}] ⇌$
$\underset{n = number of ions}{\underset{⏟}{x K^{+} + {[Fe {(CN)}_{6}]}^{x -}}} = (x + 1)$
$∴ i =$ van't Hoff factor $= 1 + (n - 1) α$
Where, α = degree of ionization
$= 1 = 1 + (x + 1 - 1) 1 = (x + 1)$
$Δ T_{f} = molality \times K_{f} \times i$
$7 \times 10^{- 3} = 1 \times 10^{- 3} \times 1.86 \times (x + 1)$
$x + 1 = 3.76 \times 4$
$∴ x = 3$

CHXII02:SOLUTIONS

319089 A solution of x moles of sucrose in 100 g of water freezes at $- 0 {.2}^{\circ} C$ . As ice separates the freezing point goes down to $0 .2 5^{^{\circ}} C$ . How many grams of ice would have separated?

1 18 g

2 20 g

3 25 g

4 23 g

Explanation:

${(m o l a l i t y)}_{i} = \frac{x \times 1000}{100} = 10 x = 0 .2 \times K_{b}$
${(m o l a l i t y)}_{f} = \frac{x \times 1000}{w} = 0 .25 \times K_{b}$
$S o \frac{0 .2}{0 .25} = \frac{w}{100}$
$\Rightarrow S o, w = 80 g$
Hence, ice separated = 20 g.

CHXII02:SOLUTIONS

319090 Assertion :
Molecular mass of polymers cannot be calculated by using boiling point or freezing point method.
Reason :
Polymer solutions possess a low boiling point or freezing point.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

CHXII02:SOLUTIONS

319091 One molal aqueous solution of urea freezes at $- {1.86}^{\circ} C$ . Aqueous solution of $C_{n} H_{2 n} O_{n}$ $36 %$ by mass of water freezes at $- {3.72}^{\circ} C$ . What is the value of n ?

1 3

2 6

3 12

4 5

Explanation:

Molality of urea solution $= 1 mol {kg}^{- 1}$
$Δ T_{f} (urea) = 0 - (- {1.86}^{\circ}) = {1.86}^{\circ}$
$Δ T_{f} = (\frac{1000 W_{2}}{M_{2} W_{1}}) K_{f} = molality K_{f}$
$∴ {1.86}^{\circ} = K_{f} \times 1 mo l^{- 1} kg$
$∴ K_{f} = 186^{\circ} mo l^{- 1} kg$
$C_{n} H_{2n} O_{n} = 36 g in 100 g H_{2}$
$Δ T_{f} = (C_{n} H_{2n} O_{n}) = \frac{1000 K_{f} W_{2}}{M_{2} W_{1}}$
$∴ M_{2} (C_{n} H_{2n} O_{n}) = \frac{1000 K_{f} W_{2}}{Δ T_{f} W_{1}}$
$= \frac{1000 \times 1.86 \times 36}{3.72 \times 100}$
$M_{2} (C_{n} H_{2n} O_{n}) = 180 g mo l^{- 1}$
$12 n + 2 n + 16 n = 180$
$∴ 30 n = 180$
$n = 6$

CHXII02:SOLUTIONS

319088 The freezing point depression of a $1.00 \times 10^{- 3}$ molal aqueous aqueous solution of $K_{x} [F e {(C N)}_{6}]$ is $7.10 \times 10^{- 3} K .$ If $K_{f} (H_{2} O)$ $= 1.86 K K g m o l^{- 1},$ what is the value of x ?

1 5

2 8

3 3

4 4

Explanation:

$K_{x} [Fe {(CN)}_{6}] ⇌$
$\underset{n = number of ions}{\underset{⏟}{x K^{+} + {[Fe {(CN)}_{6}]}^{x -}}} = (x + 1)$
$∴ i =$ van't Hoff factor $= 1 + (n - 1) α$
Where, α = degree of ionization
$= 1 = 1 + (x + 1 - 1) 1 = (x + 1)$
$Δ T_{f} = molality \times K_{f} \times i$
$7 \times 10^{- 3} = 1 \times 10^{- 3} \times 1.86 \times (x + 1)$
$x + 1 = 3.76 \times 4$
$∴ x = 3$

CHXII02:SOLUTIONS

319089 A solution of x moles of sucrose in 100 g of water freezes at $- 0 {.2}^{\circ} C$ . As ice separates the freezing point goes down to $0 .2 5^{^{\circ}} C$ . How many grams of ice would have separated?

1 18 g

2 20 g

3 25 g

4 23 g

Explanation:

${(m o l a l i t y)}_{i} = \frac{x \times 1000}{100} = 10 x = 0 .2 \times K_{b}$
${(m o l a l i t y)}_{f} = \frac{x \times 1000}{w} = 0 .25 \times K_{b}$
$S o \frac{0 .2}{0 .25} = \frac{w}{100}$
$\Rightarrow S o, w = 80 g$
Hence, ice separated = 20 g.

CHXII02:SOLUTIONS

319090 Assertion :
Molecular mass of polymers cannot be calculated by using boiling point or freezing point method.
Reason :
Polymer solutions possess a low boiling point or freezing point.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

CHXII02:SOLUTIONS

319091 One molal aqueous solution of urea freezes at $- {1.86}^{\circ} C$ . Aqueous solution of $C_{n} H_{2 n} O_{n}$ $36 %$ by mass of water freezes at $- {3.72}^{\circ} C$ . What is the value of n ?

1 3

2 6

3 12

4 5

Explanation:

Molality of urea solution $= 1 mol {kg}^{- 1}$
$Δ T_{f} (urea) = 0 - (- {1.86}^{\circ}) = {1.86}^{\circ}$
$Δ T_{f} = (\frac{1000 W_{2}}{M_{2} W_{1}}) K_{f} = molality K_{f}$
$∴ {1.86}^{\circ} = K_{f} \times 1 mo l^{- 1} kg$
$∴ K_{f} = 186^{\circ} mo l^{- 1} kg$
$C_{n} H_{2n} O_{n} = 36 g in 100 g H_{2}$
$Δ T_{f} = (C_{n} H_{2n} O_{n}) = \frac{1000 K_{f} W_{2}}{M_{2} W_{1}}$
$∴ M_{2} (C_{n} H_{2n} O_{n}) = \frac{1000 K_{f} W_{2}}{Δ T_{f} W_{1}}$
$= \frac{1000 \times 1.86 \times 36}{3.72 \times 100}$
$M_{2} (C_{n} H_{2n} O_{n}) = 180 g mo l^{- 1}$
$12 n + 2 n + 16 n = 180$
$∴ 30 n = 180$
$n = 6$

CHXII02:SOLUTIONS

319088 The freezing point depression of a $1.00 \times 10^{- 3}$ molal aqueous aqueous solution of $K_{x} [F e {(C N)}_{6}]$ is $7.10 \times 10^{- 3} K .$ If $K_{f} (H_{2} O)$ $= 1.86 K K g m o l^{- 1},$ what is the value of x ?

1 5

2 8

3 3

4 4

Explanation:

$K_{x} [Fe {(CN)}_{6}] ⇌$
$\underset{n = number of ions}{\underset{⏟}{x K^{+} + {[Fe {(CN)}_{6}]}^{x -}}} = (x + 1)$
$∴ i =$ van't Hoff factor $= 1 + (n - 1) α$
Where, α = degree of ionization
$= 1 = 1 + (x + 1 - 1) 1 = (x + 1)$
$Δ T_{f} = molality \times K_{f} \times i$
$7 \times 10^{- 3} = 1 \times 10^{- 3} \times 1.86 \times (x + 1)$
$x + 1 = 3.76 \times 4$
$∴ x = 3$

CHXII02:SOLUTIONS

319089 A solution of x moles of sucrose in 100 g of water freezes at $- 0 {.2}^{\circ} C$ . As ice separates the freezing point goes down to $0 .2 5^{^{\circ}} C$ . How many grams of ice would have separated?

1 18 g

2 20 g

3 25 g

4 23 g

Explanation:

${(m o l a l i t y)}_{i} = \frac{x \times 1000}{100} = 10 x = 0 .2 \times K_{b}$
${(m o l a l i t y)}_{f} = \frac{x \times 1000}{w} = 0 .25 \times K_{b}$
$S o \frac{0 .2}{0 .25} = \frac{w}{100}$
$\Rightarrow S o, w = 80 g$
Hence, ice separated = 20 g.

CHXII02:SOLUTIONS

319090 Assertion :
Molecular mass of polymers cannot be calculated by using boiling point or freezing point method.
Reason :
Polymer solutions possess a low boiling point or freezing point.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

CHXII02:SOLUTIONS

319091 One molal aqueous solution of urea freezes at $- {1.86}^{\circ} C$ . Aqueous solution of $C_{n} H_{2 n} O_{n}$ $36 %$ by mass of water freezes at $- {3.72}^{\circ} C$ . What is the value of n ?

1 3

2 6

3 12

4 5

Explanation:

Molality of urea solution $= 1 mol {kg}^{- 1}$
$Δ T_{f} (urea) = 0 - (- {1.86}^{\circ}) = {1.86}^{\circ}$
$Δ T_{f} = (\frac{1000 W_{2}}{M_{2} W_{1}}) K_{f} = molality K_{f}$
$∴ {1.86}^{\circ} = K_{f} \times 1 mo l^{- 1} kg$
$∴ K_{f} = 186^{\circ} mo l^{- 1} kg$
$C_{n} H_{2n} O_{n} = 36 g in 100 g H_{2}$
$Δ T_{f} = (C_{n} H_{2n} O_{n}) = \frac{1000 K_{f} W_{2}}{M_{2} W_{1}}$
$∴ M_{2} (C_{n} H_{2n} O_{n}) = \frac{1000 K_{f} W_{2}}{Δ T_{f} W_{1}}$
$= \frac{1000 \times 1.86 \times 36}{3.72 \times 100}$
$M_{2} (C_{n} H_{2n} O_{n}) = 180 g mo l^{- 1}$
$12 n + 2 n + 16 n = 180$
$∴ 30 n = 180$
$n = 6$

CHXII02:SOLUTIONS

319088 The freezing point depression of a $1.00 \times 10^{- 3}$ molal aqueous aqueous solution of $K_{x} [F e {(C N)}_{6}]$ is $7.10 \times 10^{- 3} K .$ If $K_{f} (H_{2} O)$ $= 1.86 K K g m o l^{- 1},$ what is the value of x ?

1 5

2 8

3 3

4 4

Explanation:

$K_{x} [Fe {(CN)}_{6}] ⇌$
$\underset{n = number of ions}{\underset{⏟}{x K^{+} + {[Fe {(CN)}_{6}]}^{x -}}} = (x + 1)$
$∴ i =$ van't Hoff factor $= 1 + (n - 1) α$
Where, α = degree of ionization
$= 1 = 1 + (x + 1 - 1) 1 = (x + 1)$
$Δ T_{f} = molality \times K_{f} \times i$
$7 \times 10^{- 3} = 1 \times 10^{- 3} \times 1.86 \times (x + 1)$
$x + 1 = 3.76 \times 4$
$∴ x = 3$

CHXII02:SOLUTIONS

319089 A solution of x moles of sucrose in 100 g of water freezes at $- 0 {.2}^{\circ} C$ . As ice separates the freezing point goes down to $0 .2 5^{^{\circ}} C$ . How many grams of ice would have separated?

1 18 g

2 20 g

3 25 g

4 23 g

Explanation:

${(m o l a l i t y)}_{i} = \frac{x \times 1000}{100} = 10 x = 0 .2 \times K_{b}$
${(m o l a l i t y)}_{f} = \frac{x \times 1000}{w} = 0 .25 \times K_{b}$
$S o \frac{0 .2}{0 .25} = \frac{w}{100}$
$\Rightarrow S o, w = 80 g$
Hence, ice separated = 20 g.

CHXII02:SOLUTIONS

319090 Assertion :
Molecular mass of polymers cannot be calculated by using boiling point or freezing point method.
Reason :
Polymer solutions possess a low boiling point or freezing point.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

CHXII02:SOLUTIONS

319091 One molal aqueous solution of urea freezes at $- {1.86}^{\circ} C$ . Aqueous solution of $C_{n} H_{2 n} O_{n}$ $36 %$ by mass of water freezes at $- {3.72}^{\circ} C$ . What is the value of n ?

1 3

2 6

3 12

4 5

Explanation:

Molality of urea solution $= 1 mol {kg}^{- 1}$
$Δ T_{f} (urea) = 0 - (- {1.86}^{\circ}) = {1.86}^{\circ}$
$Δ T_{f} = (\frac{1000 W_{2}}{M_{2} W_{1}}) K_{f} = molality K_{f}$
$∴ {1.86}^{\circ} = K_{f} \times 1 mo l^{- 1} kg$
$∴ K_{f} = 186^{\circ} mo l^{- 1} kg$
$C_{n} H_{2n} O_{n} = 36 g in 100 g H_{2}$
$Δ T_{f} = (C_{n} H_{2n} O_{n}) = \frac{1000 K_{f} W_{2}}{M_{2} W_{1}}$
$∴ M_{2} (C_{n} H_{2n} O_{n}) = \frac{1000 K_{f} W_{2}}{Δ T_{f} W_{1}}$
$= \frac{1000 \times 1.86 \times 36}{3.72 \times 100}$
$M_{2} (C_{n} H_{2n} O_{n}) = 180 g mo l^{- 1}$
$12 n + 2 n + 16 n = 180$
$∴ 30 n = 180$
$n = 6$

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