QBManager

SOLUTIONS

277353 A solution of 1.25 of ' $P$ ' in $50 g$ of water lowers freezing point by ${0.3}^{\circ} C$ . Molar mass of ' $P$ ' is 94 . $K_{f (water)} = 1.86 K kg {mol}^{- 1}$ . The degree of association of ' $P$ ' in water is

1

80 %

2

60 %

3

65 %

4

75 %

Explanation:

From formula
$\begin{aligned} Δ T_{f} & = i \times K_{f} \times m \\ = i \times K_{f} \times \frac{W_{1} \times 1000}{W_{1} \times M_{1}} \\ 0.3 & = i \times 1.86 \times \frac{1.25 \times 1000}{50 \times 94} \\ i & = 0.6064 \end{aligned}$
now, degree of association of ' $P$ ' in water $α = \frac{i - 1}{\frac{1}{n} - 1}$
$\begin{aligned} = \frac{0.6064 - 1}{\frac{1}{2} - 1} \\ = 0.7872 \\ = 78.72 % \\ ≃ 80 % \end{aligned}$

Karnataka-CET-2014

SOLUTIONS

277354 An aqueous solution containing $3 g$ of a solute of molar mass $111.6 g {mol}^{- 1}$ in a certain mass of water freezes at $- 0.125^{\circ} C$ . The mass of water in grams present in the solution is $(K_{f} = 1.86 K kg$ mol $^{- 1}$ )

1 300

2 600

3 500

4 400

5 250

Explanation:

From formula-
$Δ T_{f} = \frac{K_{f} \times W \times 1000}{M \times W} {\begin{cases} w and M = weight and molar \\ mass of solute \\ W = weight of solvent or water \end{cases}}$
Putting the value,
$Δ T_{f} = 0 - (- 0.125)$
$= 0.125$
$0.125 = \frac{1.86 \times 3 \times 1000}{111.6 \times W}$
$W = \frac{1.86 \times 3 \times 1000}{0.125 \times 111.6}$
$= 400 g$

Kerala-CEE-2014

SOLUTIONS

277356 A 0.001 molal solution of $[Pt {({NH}_{3})}_{4} {Cl}_{4}]$ in water had a freezing point depression of ${0.0054}^{0} C$ . If $K_{f}$ for water is 1.80 the correct formulation of the above molecule is

1

[Pt {({NH}_{3})}_{4} {Cl}_{3}] Cl

2

[Pt {({NH}_{3})}_{4} {Cl}_{2}] {Cl}_{2}

3

[Pt {({NH}_{3})}_{4} Cl] {Cl}_{2}

4

[Pt {({NH}_{3})}_{4} {Cl}_{4}]

Explanation:

According to the depression in freezing point Given that,
$\begin{matrix} Δ T_{f} = 1.80, m = 0.001 \\ i = \frac{0.0054}{0.001 \times 1.80} = 3 \\ [Pt {({HN}_{3})}_{4} {Cl}_{2}] {Cl}_{2} \overset{water}{⇌} {[Pt {({NH}_{3})}_{4} {Cl}_{2}]}^{2 +} + 2 {Cl}^{-} \end{matrix}$

UPTU/UPSEE-2014

SOLUTIONS

277358 Vapour pressure in $mm Hg$ of 0.1 mole of urea in $180 g$ of water at $25^{\circ} C$ is (The vapour pressure of water at $25^{\circ} C$ is $24 m m H g)$

1 2.376

2 20.76

3 23.76

4 24.76

Explanation:

Given that, $P^{\circ} = 24, M = 18, W = 180$
Number of moles urea $= \frac{W}{m} = 0.1$
According to the Raoult's law of dilute solution -
$\begin{aligned} \frac{P^{\circ} - P_{s}}{P^{\circ}} = \frac{W}{m} \cdot \frac{M}{W} \\ So, \frac{24 - P_{s}}{24} = 0.1 \times \frac{18}{180} \\ 24 - P_{s} = 0.01 \times 24 \\ P_{s} = 24 - 0.24 = 23.76 mm Hg \end{aligned}$

AP-EAMCET (Engg.)-2014

SOLUTIONS

277353 A solution of 1.25 of ' $P$ ' in $50 g$ of water lowers freezing point by ${0.3}^{\circ} C$ . Molar mass of ' $P$ ' is 94 . $K_{f (water)} = 1.86 K kg {mol}^{- 1}$ . The degree of association of ' $P$ ' in water is

1

80 %

2

60 %

3

65 %

4

75 %

Explanation:

From formula
$\begin{aligned} Δ T_{f} & = i \times K_{f} \times m \\ = i \times K_{f} \times \frac{W_{1} \times 1000}{W_{1} \times M_{1}} \\ 0.3 & = i \times 1.86 \times \frac{1.25 \times 1000}{50 \times 94} \\ i & = 0.6064 \end{aligned}$
now, degree of association of ' $P$ ' in water $α = \frac{i - 1}{\frac{1}{n} - 1}$
$\begin{aligned} = \frac{0.6064 - 1}{\frac{1}{2} - 1} \\ = 0.7872 \\ = 78.72 % \\ ≃ 80 % \end{aligned}$

Karnataka-CET-2014

SOLUTIONS

277354 An aqueous solution containing $3 g$ of a solute of molar mass $111.6 g {mol}^{- 1}$ in a certain mass of water freezes at $- 0.125^{\circ} C$ . The mass of water in grams present in the solution is $(K_{f} = 1.86 K kg$ mol $^{- 1}$ )

1 300

2 600

3 500

4 400

5 250

Explanation:

From formula-
$Δ T_{f} = \frac{K_{f} \times W \times 1000}{M \times W} {\begin{cases} w and M = weight and molar \\ mass of solute \\ W = weight of solvent or water \end{cases}}$
Putting the value,
$Δ T_{f} = 0 - (- 0.125)$
$= 0.125$
$0.125 = \frac{1.86 \times 3 \times 1000}{111.6 \times W}$
$W = \frac{1.86 \times 3 \times 1000}{0.125 \times 111.6}$
$= 400 g$

Kerala-CEE-2014

SOLUTIONS

277356 A 0.001 molal solution of $[Pt {({NH}_{3})}_{4} {Cl}_{4}]$ in water had a freezing point depression of ${0.0054}^{0} C$ . If $K_{f}$ for water is 1.80 the correct formulation of the above molecule is

1

[Pt {({NH}_{3})}_{4} {Cl}_{3}] Cl

2

[Pt {({NH}_{3})}_{4} {Cl}_{2}] {Cl}_{2}

3

[Pt {({NH}_{3})}_{4} Cl] {Cl}_{2}

4

[Pt {({NH}_{3})}_{4} {Cl}_{4}]

Explanation:

According to the depression in freezing point Given that,
$\begin{matrix} Δ T_{f} = 1.80, m = 0.001 \\ i = \frac{0.0054}{0.001 \times 1.80} = 3 \\ [Pt {({HN}_{3})}_{4} {Cl}_{2}] {Cl}_{2} \overset{water}{⇌} {[Pt {({NH}_{3})}_{4} {Cl}_{2}]}^{2 +} + 2 {Cl}^{-} \end{matrix}$

UPTU/UPSEE-2014

SOLUTIONS

277358 Vapour pressure in $mm Hg$ of 0.1 mole of urea in $180 g$ of water at $25^{\circ} C$ is (The vapour pressure of water at $25^{\circ} C$ is $24 m m H g)$

1 2.376

2 20.76

3 23.76

4 24.76

Explanation:

Given that, $P^{\circ} = 24, M = 18, W = 180$
Number of moles urea $= \frac{W}{m} = 0.1$
According to the Raoult's law of dilute solution -
$\begin{aligned} \frac{P^{\circ} - P_{s}}{P^{\circ}} = \frac{W}{m} \cdot \frac{M}{W} \\ So, \frac{24 - P_{s}}{24} = 0.1 \times \frac{18}{180} \\ 24 - P_{s} = 0.01 \times 24 \\ P_{s} = 24 - 0.24 = 23.76 mm Hg \end{aligned}$

AP-EAMCET (Engg.)-2014

SOLUTIONS

277353 A solution of 1.25 of ' $P$ ' in $50 g$ of water lowers freezing point by ${0.3}^{\circ} C$ . Molar mass of ' $P$ ' is 94 . $K_{f (water)} = 1.86 K kg {mol}^{- 1}$ . The degree of association of ' $P$ ' in water is

1

80 %

2

60 %

3

65 %

4

75 %

Explanation:

From formula
$\begin{aligned} Δ T_{f} & = i \times K_{f} \times m \\ = i \times K_{f} \times \frac{W_{1} \times 1000}{W_{1} \times M_{1}} \\ 0.3 & = i \times 1.86 \times \frac{1.25 \times 1000}{50 \times 94} \\ i & = 0.6064 \end{aligned}$
now, degree of association of ' $P$ ' in water $α = \frac{i - 1}{\frac{1}{n} - 1}$
$\begin{aligned} = \frac{0.6064 - 1}{\frac{1}{2} - 1} \\ = 0.7872 \\ = 78.72 % \\ ≃ 80 % \end{aligned}$

Karnataka-CET-2014

SOLUTIONS

277354 An aqueous solution containing $3 g$ of a solute of molar mass $111.6 g {mol}^{- 1}$ in a certain mass of water freezes at $- 0.125^{\circ} C$ . The mass of water in grams present in the solution is $(K_{f} = 1.86 K kg$ mol $^{- 1}$ )

1 300

2 600

3 500

4 400

5 250

Explanation:

From formula-
$Δ T_{f} = \frac{K_{f} \times W \times 1000}{M \times W} {\begin{cases} w and M = weight and molar \\ mass of solute \\ W = weight of solvent or water \end{cases}}$
Putting the value,
$Δ T_{f} = 0 - (- 0.125)$
$= 0.125$
$0.125 = \frac{1.86 \times 3 \times 1000}{111.6 \times W}$
$W = \frac{1.86 \times 3 \times 1000}{0.125 \times 111.6}$
$= 400 g$

Kerala-CEE-2014

SOLUTIONS

277356 A 0.001 molal solution of $[Pt {({NH}_{3})}_{4} {Cl}_{4}]$ in water had a freezing point depression of ${0.0054}^{0} C$ . If $K_{f}$ for water is 1.80 the correct formulation of the above molecule is

1

[Pt {({NH}_{3})}_{4} {Cl}_{3}] Cl

2

[Pt {({NH}_{3})}_{4} {Cl}_{2}] {Cl}_{2}

3

[Pt {({NH}_{3})}_{4} Cl] {Cl}_{2}

4

[Pt {({NH}_{3})}_{4} {Cl}_{4}]

Explanation:

According to the depression in freezing point Given that,
$\begin{matrix} Δ T_{f} = 1.80, m = 0.001 \\ i = \frac{0.0054}{0.001 \times 1.80} = 3 \\ [Pt {({HN}_{3})}_{4} {Cl}_{2}] {Cl}_{2} \overset{water}{⇌} {[Pt {({NH}_{3})}_{4} {Cl}_{2}]}^{2 +} + 2 {Cl}^{-} \end{matrix}$

UPTU/UPSEE-2014

SOLUTIONS

277358 Vapour pressure in $mm Hg$ of 0.1 mole of urea in $180 g$ of water at $25^{\circ} C$ is (The vapour pressure of water at $25^{\circ} C$ is $24 m m H g)$

1 2.376

2 20.76

3 23.76

4 24.76

Explanation:

Given that, $P^{\circ} = 24, M = 18, W = 180$
Number of moles urea $= \frac{W}{m} = 0.1$
According to the Raoult's law of dilute solution -
$\begin{aligned} \frac{P^{\circ} - P_{s}}{P^{\circ}} = \frac{W}{m} \cdot \frac{M}{W} \\ So, \frac{24 - P_{s}}{24} = 0.1 \times \frac{18}{180} \\ 24 - P_{s} = 0.01 \times 24 \\ P_{s} = 24 - 0.24 = 23.76 mm Hg \end{aligned}$

AP-EAMCET (Engg.)-2014

SOLUTIONS

277353 A solution of 1.25 of ' $P$ ' in $50 g$ of water lowers freezing point by ${0.3}^{\circ} C$ . Molar mass of ' $P$ ' is 94 . $K_{f (water)} = 1.86 K kg {mol}^{- 1}$ . The degree of association of ' $P$ ' in water is

1

80 %

2

60 %

3

65 %

4

75 %

Explanation:

From formula
$\begin{aligned} Δ T_{f} & = i \times K_{f} \times m \\ = i \times K_{f} \times \frac{W_{1} \times 1000}{W_{1} \times M_{1}} \\ 0.3 & = i \times 1.86 \times \frac{1.25 \times 1000}{50 \times 94} \\ i & = 0.6064 \end{aligned}$
now, degree of association of ' $P$ ' in water $α = \frac{i - 1}{\frac{1}{n} - 1}$
$\begin{aligned} = \frac{0.6064 - 1}{\frac{1}{2} - 1} \\ = 0.7872 \\ = 78.72 % \\ ≃ 80 % \end{aligned}$

Karnataka-CET-2014

SOLUTIONS

277354 An aqueous solution containing $3 g$ of a solute of molar mass $111.6 g {mol}^{- 1}$ in a certain mass of water freezes at $- 0.125^{\circ} C$ . The mass of water in grams present in the solution is $(K_{f} = 1.86 K kg$ mol $^{- 1}$ )

1 300

2 600

3 500

4 400

5 250

Explanation:

From formula-
$Δ T_{f} = \frac{K_{f} \times W \times 1000}{M \times W} {\begin{cases} w and M = weight and molar \\ mass of solute \\ W = weight of solvent or water \end{cases}}$
Putting the value,
$Δ T_{f} = 0 - (- 0.125)$
$= 0.125$
$0.125 = \frac{1.86 \times 3 \times 1000}{111.6 \times W}$
$W = \frac{1.86 \times 3 \times 1000}{0.125 \times 111.6}$
$= 400 g$

Kerala-CEE-2014

SOLUTIONS

277356 A 0.001 molal solution of $[Pt {({NH}_{3})}_{4} {Cl}_{4}]$ in water had a freezing point depression of ${0.0054}^{0} C$ . If $K_{f}$ for water is 1.80 the correct formulation of the above molecule is

1

[Pt {({NH}_{3})}_{4} {Cl}_{3}] Cl

2

[Pt {({NH}_{3})}_{4} {Cl}_{2}] {Cl}_{2}

3

[Pt {({NH}_{3})}_{4} Cl] {Cl}_{2}

4

[Pt {({NH}_{3})}_{4} {Cl}_{4}]

Explanation:

According to the depression in freezing point Given that,
$\begin{matrix} Δ T_{f} = 1.80, m = 0.001 \\ i = \frac{0.0054}{0.001 \times 1.80} = 3 \\ [Pt {({HN}_{3})}_{4} {Cl}_{2}] {Cl}_{2} \overset{water}{⇌} {[Pt {({NH}_{3})}_{4} {Cl}_{2}]}^{2 +} + 2 {Cl}^{-} \end{matrix}$

UPTU/UPSEE-2014

SOLUTIONS

277358 Vapour pressure in $mm Hg$ of 0.1 mole of urea in $180 g$ of water at $25^{\circ} C$ is (The vapour pressure of water at $25^{\circ} C$ is $24 m m H g)$

1 2.376

2 20.76

3 23.76

4 24.76

Explanation:

Given that, $P^{\circ} = 24, M = 18, W = 180$
Number of moles urea $= \frac{W}{m} = 0.1$
According to the Raoult's law of dilute solution -
$\begin{aligned} \frac{P^{\circ} - P_{s}}{P^{\circ}} = \frac{W}{m} \cdot \frac{M}{W} \\ So, \frac{24 - P_{s}}{24} = 0.1 \times \frac{18}{180} \\ 24 - P_{s} = 0.01 \times 24 \\ P_{s} = 24 - 0.24 = 23.76 mm Hg \end{aligned}$

AP-EAMCET (Engg.)-2014

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation: