QBManager

CHXII02:SOLUTIONS

319145 A solution of urea (molar mass 60) boils at $100 .1 8^{^{\circ}} C$ at atmospheric pressure. If $K_{f} a n d K_{b}$ for water are $1 .86 a n d 0 .512 K m o l a l i t y^{- 1}$ respectively, the above solution will freeze at

1

- 0 .65 4^{^{\circ}} C

2

0 .65 4^{^{\circ}} C

3

- 6 .5 4^{^{\circ}} C

4

6 .5 4^{^{\circ}} C

Explanation:

$\frac{Δ T_{b}}{Δ T_{f}} = \frac{K_{b}}{K_{f}}$
$∴ Δ T_{f} = \frac{K_{f}}{K_{b}} \times Δ T_{b} = \frac{1 .86}{0 .512} \times 0 .18 = 0 .654$
$∴ F . P . = 0 - 0 .654 = - 0 .65 4^{^{\circ}} C$

CHXII02:SOLUTIONS

319146 Equimolal solutions of two non - electrolytes in the same solvent have:

1 different b.pt. and different f.pt.

2 same b.pt. and same f.pt.

3 same b.pt. but different f.pt.

4 different b.pt. but same f.pt.

Explanation:

$Δ T = K_{f} \times m o l a l i t y,$
$K_{f}$ is common for both thus, same $Δ T_{f}$
$Δ T_{b} = K_{b} \times m o l a l i t y,$
$K_{b}$ is common for both thus, same $Δ T_{b}$ .

CHXII02:SOLUTIONS

319147 The molal elevation of an unknown solution is equal to the molal elevation of 0.1 M solution of urea. The concentration of unknown solution is

1 1 M

2 0.01 M

3 0.1 M

4 None of these

Explanation:

${(Δ T_{b})}_{x} = {(Δ T_{b})}_{u r e a}$
$K_{b} m_{x} = K_{b} \times 0.1$
$m_{x} = 0.1$
(Considering molarity $=$ Molality)

CHXII02:SOLUTIONS

319148 The unit of ebullioscopic constant is

1

K k g m o l^{- 1} o r K {(m o l a l i t y)}^{- 1}

2

m o l k g K^{- 1} o r K^{- 1} (m o l a l i t y)

3

k g m o l^{- 1} K^{- 1} o r K^{- 1} {(m o l a l i t y)}^{- 1}

4

K m o l k g^{- 1} o r K (m o l a l i t y)

Explanation:

As we know from elevation in boiling point that
$Δ T_{b} = K_{b} m$
$K_{b} = \frac{Δ T_{b}}{m}$
$u n i t o f K_{b} = \frac{u n i t o f Δ T_{b}}{u n i t o f m} = \frac{K}{m o l a l i t y}$
$= \frac{K}{m o l k g^{- 1}} = K m o l^{- 1} k g$

NCERT Exampler

CHXII02:SOLUTIONS

319145 A solution of urea (molar mass 60) boils at $100 .1 8^{^{\circ}} C$ at atmospheric pressure. If $K_{f} a n d K_{b}$ for water are $1 .86 a n d 0 .512 K m o l a l i t y^{- 1}$ respectively, the above solution will freeze at

1

- 0 .65 4^{^{\circ}} C

2

0 .65 4^{^{\circ}} C

3

- 6 .5 4^{^{\circ}} C

4

6 .5 4^{^{\circ}} C

Explanation:

$\frac{Δ T_{b}}{Δ T_{f}} = \frac{K_{b}}{K_{f}}$
$∴ Δ T_{f} = \frac{K_{f}}{K_{b}} \times Δ T_{b} = \frac{1 .86}{0 .512} \times 0 .18 = 0 .654$
$∴ F . P . = 0 - 0 .654 = - 0 .65 4^{^{\circ}} C$

CHXII02:SOLUTIONS

319146 Equimolal solutions of two non - electrolytes in the same solvent have:

1 different b.pt. and different f.pt.

2 same b.pt. and same f.pt.

3 same b.pt. but different f.pt.

4 different b.pt. but same f.pt.

Explanation:

$Δ T = K_{f} \times m o l a l i t y,$
$K_{f}$ is common for both thus, same $Δ T_{f}$
$Δ T_{b} = K_{b} \times m o l a l i t y,$
$K_{b}$ is common for both thus, same $Δ T_{b}$ .

CHXII02:SOLUTIONS

319147 The molal elevation of an unknown solution is equal to the molal elevation of 0.1 M solution of urea. The concentration of unknown solution is

1 1 M

2 0.01 M

3 0.1 M

4 None of these

Explanation:

${(Δ T_{b})}_{x} = {(Δ T_{b})}_{u r e a}$
$K_{b} m_{x} = K_{b} \times 0.1$
$m_{x} = 0.1$
(Considering molarity $=$ Molality)

CHXII02:SOLUTIONS

319148 The unit of ebullioscopic constant is

1

K k g m o l^{- 1} o r K {(m o l a l i t y)}^{- 1}

2

m o l k g K^{- 1} o r K^{- 1} (m o l a l i t y)

3

k g m o l^{- 1} K^{- 1} o r K^{- 1} {(m o l a l i t y)}^{- 1}

4

K m o l k g^{- 1} o r K (m o l a l i t y)

Explanation:

As we know from elevation in boiling point that
$Δ T_{b} = K_{b} m$
$K_{b} = \frac{Δ T_{b}}{m}$
$u n i t o f K_{b} = \frac{u n i t o f Δ T_{b}}{u n i t o f m} = \frac{K}{m o l a l i t y}$
$= \frac{K}{m o l k g^{- 1}} = K m o l^{- 1} k g$

NCERT Exampler

CHXII02:SOLUTIONS

319145 A solution of urea (molar mass 60) boils at $100 .1 8^{^{\circ}} C$ at atmospheric pressure. If $K_{f} a n d K_{b}$ for water are $1 .86 a n d 0 .512 K m o l a l i t y^{- 1}$ respectively, the above solution will freeze at

1

- 0 .65 4^{^{\circ}} C

2

0 .65 4^{^{\circ}} C

3

- 6 .5 4^{^{\circ}} C

4

6 .5 4^{^{\circ}} C

Explanation:

$\frac{Δ T_{b}}{Δ T_{f}} = \frac{K_{b}}{K_{f}}$
$∴ Δ T_{f} = \frac{K_{f}}{K_{b}} \times Δ T_{b} = \frac{1 .86}{0 .512} \times 0 .18 = 0 .654$
$∴ F . P . = 0 - 0 .654 = - 0 .65 4^{^{\circ}} C$

CHXII02:SOLUTIONS

319146 Equimolal solutions of two non - electrolytes in the same solvent have:

1 different b.pt. and different f.pt.

2 same b.pt. and same f.pt.

3 same b.pt. but different f.pt.

4 different b.pt. but same f.pt.

Explanation:

$Δ T = K_{f} \times m o l a l i t y,$
$K_{f}$ is common for both thus, same $Δ T_{f}$
$Δ T_{b} = K_{b} \times m o l a l i t y,$
$K_{b}$ is common for both thus, same $Δ T_{b}$ .

CHXII02:SOLUTIONS

319147 The molal elevation of an unknown solution is equal to the molal elevation of 0.1 M solution of urea. The concentration of unknown solution is

1 1 M

2 0.01 M

3 0.1 M

4 None of these

Explanation:

${(Δ T_{b})}_{x} = {(Δ T_{b})}_{u r e a}$
$K_{b} m_{x} = K_{b} \times 0.1$
$m_{x} = 0.1$
(Considering molarity $=$ Molality)

CHXII02:SOLUTIONS

319148 The unit of ebullioscopic constant is

1

K k g m o l^{- 1} o r K {(m o l a l i t y)}^{- 1}

2

m o l k g K^{- 1} o r K^{- 1} (m o l a l i t y)

3

k g m o l^{- 1} K^{- 1} o r K^{- 1} {(m o l a l i t y)}^{- 1}

4

K m o l k g^{- 1} o r K (m o l a l i t y)

Explanation:

As we know from elevation in boiling point that
$Δ T_{b} = K_{b} m$
$K_{b} = \frac{Δ T_{b}}{m}$
$u n i t o f K_{b} = \frac{u n i t o f Δ T_{b}}{u n i t o f m} = \frac{K}{m o l a l i t y}$
$= \frac{K}{m o l k g^{- 1}} = K m o l^{- 1} k g$

NCERT Exampler

CHXII02:SOLUTIONS

319145 A solution of urea (molar mass 60) boils at $100 .1 8^{^{\circ}} C$ at atmospheric pressure. If $K_{f} a n d K_{b}$ for water are $1 .86 a n d 0 .512 K m o l a l i t y^{- 1}$ respectively, the above solution will freeze at

1

- 0 .65 4^{^{\circ}} C

2

0 .65 4^{^{\circ}} C

3

- 6 .5 4^{^{\circ}} C

4

6 .5 4^{^{\circ}} C

Explanation:

$\frac{Δ T_{b}}{Δ T_{f}} = \frac{K_{b}}{K_{f}}$
$∴ Δ T_{f} = \frac{K_{f}}{K_{b}} \times Δ T_{b} = \frac{1 .86}{0 .512} \times 0 .18 = 0 .654$
$∴ F . P . = 0 - 0 .654 = - 0 .65 4^{^{\circ}} C$

CHXII02:SOLUTIONS

319146 Equimolal solutions of two non - electrolytes in the same solvent have:

1 different b.pt. and different f.pt.

2 same b.pt. and same f.pt.

3 same b.pt. but different f.pt.

4 different b.pt. but same f.pt.

Explanation:

$Δ T = K_{f} \times m o l a l i t y,$
$K_{f}$ is common for both thus, same $Δ T_{f}$
$Δ T_{b} = K_{b} \times m o l a l i t y,$
$K_{b}$ is common for both thus, same $Δ T_{b}$ .

CHXII02:SOLUTIONS

319147 The molal elevation of an unknown solution is equal to the molal elevation of 0.1 M solution of urea. The concentration of unknown solution is

1 1 M

2 0.01 M

3 0.1 M

4 None of these

Explanation:

${(Δ T_{b})}_{x} = {(Δ T_{b})}_{u r e a}$
$K_{b} m_{x} = K_{b} \times 0.1$
$m_{x} = 0.1$
(Considering molarity $=$ Molality)

CHXII02:SOLUTIONS

319148 The unit of ebullioscopic constant is

1

K k g m o l^{- 1} o r K {(m o l a l i t y)}^{- 1}

2

m o l k g K^{- 1} o r K^{- 1} (m o l a l i t y)

3

k g m o l^{- 1} K^{- 1} o r K^{- 1} {(m o l a l i t y)}^{- 1}

4

K m o l k g^{- 1} o r K (m o l a l i t y)

Explanation:

As we know from elevation in boiling point that
$Δ T_{b} = K_{b} m$
$K_{b} = \frac{Δ T_{b}}{m}$
$u n i t o f K_{b} = \frac{u n i t o f Δ T_{b}}{u n i t o f m} = \frac{K}{m o l a l i t y}$
$= \frac{K}{m o l k g^{- 1}} = K m o l^{- 1} k g$

NCERT Exampler

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation: