QBManager

4 RBTS PAPER

163825 Elevation in boiling point was ${0.52}^{\circ} C$ when $6 g$ of a compound $X$ was dissolved in $100 g$ of water. Molecular weight of $X$ is: $(Kb$ for water $= 0.52 K$ ${mol}^{- 1}$ )

1 120

2 60

3 100

4 342

Explanation:

$Δ T_{b} = K_{b} \times m$
$0.52 = 0.52 \times m$
$m = 1$
$\frac{6}{M_{X}} \times \frac{1000}{100} = 1$
$M_{x} = 60$

NCERT-XII-17

4 RBTS PAPER

163826 An aqueous solution containing $1 g$ of urea boils at ${100.25}^{\circ} C$ . The aqueous solution containing $3 g$ of glucose in the same volume will boil at -

1

{100.75}^{\circ} C

2

{100.5}^{\circ} C

3

100^{\circ} C

4

{100.25}^{\circ} C

Explanation:

For Urea solution
$Δ T_{b} = K_{b} (\frac{1}{60 \times W_{solvent}})$
$0.25 = K_{b} (\frac{1}{60 \times W_{solvent}})$

For glucose solution
$(T_{b} - 100) = K_{b} (\frac{1}{180 \times W_{solvent}})$

From $(2) \div (1)$
$\frac{T_{b} - 100}{0.25} = 1$
$T_{b} = 100 + 0.25$
$T_{b} = 100.25$

NCERT-XII-17

4 RBTS PAPER

163827 Pure benzene freezes at ${5.45}^{\circ} C$ at a certain place but a $0.374 m$ solution of tetrachloroethane in benzene freezes at ${3.55}^{\circ} C$ . The $K_{f}$ for benzene is-

1

5.08 K Kg {mol}^{- 1}

2

508 K Kg {mol}^{- 1}

3

0.508 K Kg {mol}^{- 1}

4

{50.8}^{\circ} C Kg {mol}^{- 1}

Explanation:

$Δ T_{f} = K_{f} \times m$
$T_{f}^{0} - T_{f} = K_{f} \times m$
$5.45 - 3.55 = K_{f} \times 0.374$
$K_{f} = \frac{1.9}{0.374} = 5.08 K Kg / mol$

NCERT-XII-18

4 RBTS PAPER

163828 Molal depression constant of water is $1.86 K Kg$ ${mol}^{- 1} .0 .02$ mole of urea dissolved in $100 g$ of water will produce a depression in freezing point of:

1

{0.186}^{\circ} C

2

{0.372}^{\circ} C

3

{1.86}^{\circ} C

4

{3.72}^{\circ} C

Explanation:

$Δ T_{f} = K_{f} \times m$
$Δ T_{f} = 1.86 \times \frac{0.02}{100} \times 1000$
$Δ T_{f} = 1.86 \times 0.02 \times 10$
$Δ T_{f} = {0.372}^{\circ} C$

NCERT-XII-18

4 RBTS PAPER

163830 A solution containing $8.6 g$ urea in one litre was found to be isotonic with $0.5 %$ (wt./vol) solution of an organic non volatile solute. The molecular weight of organic solute is:

1 348.9

2 34.89

3 3489

4 861.2

Explanation:

$π_{1} = π_{2}$
$C_{1} R T = C_{2} R T$
$C_{1} = C_{2}$
$\frac{8.6}{60} = \frac{0.5}{M \times 0.1}$
$M = 34.89$

NCERT-XII-21

4 RBTS PAPER

163825 Elevation in boiling point was ${0.52}^{\circ} C$ when $6 g$ of a compound $X$ was dissolved in $100 g$ of water. Molecular weight of $X$ is: $(Kb$ for water $= 0.52 K$ ${mol}^{- 1}$ )

1 120

2 60

3 100

4 342

Explanation:

$Δ T_{b} = K_{b} \times m$
$0.52 = 0.52 \times m$
$m = 1$
$\frac{6}{M_{X}} \times \frac{1000}{100} = 1$
$M_{x} = 60$

NCERT-XII-17

4 RBTS PAPER

163826 An aqueous solution containing $1 g$ of urea boils at ${100.25}^{\circ} C$ . The aqueous solution containing $3 g$ of glucose in the same volume will boil at -

1

{100.75}^{\circ} C

2

{100.5}^{\circ} C

3

100^{\circ} C

4

{100.25}^{\circ} C

Explanation:

For Urea solution
$Δ T_{b} = K_{b} (\frac{1}{60 \times W_{solvent}})$
$0.25 = K_{b} (\frac{1}{60 \times W_{solvent}})$

For glucose solution
$(T_{b} - 100) = K_{b} (\frac{1}{180 \times W_{solvent}})$

From $(2) \div (1)$
$\frac{T_{b} - 100}{0.25} = 1$
$T_{b} = 100 + 0.25$
$T_{b} = 100.25$

NCERT-XII-17

4 RBTS PAPER

163827 Pure benzene freezes at ${5.45}^{\circ} C$ at a certain place but a $0.374 m$ solution of tetrachloroethane in benzene freezes at ${3.55}^{\circ} C$ . The $K_{f}$ for benzene is-

1

5.08 K Kg {mol}^{- 1}

2

508 K Kg {mol}^{- 1}

3

0.508 K Kg {mol}^{- 1}

4

{50.8}^{\circ} C Kg {mol}^{- 1}

Explanation:

$Δ T_{f} = K_{f} \times m$
$T_{f}^{0} - T_{f} = K_{f} \times m$
$5.45 - 3.55 = K_{f} \times 0.374$
$K_{f} = \frac{1.9}{0.374} = 5.08 K Kg / mol$

NCERT-XII-18

4 RBTS PAPER

163828 Molal depression constant of water is $1.86 K Kg$ ${mol}^{- 1} .0 .02$ mole of urea dissolved in $100 g$ of water will produce a depression in freezing point of:

1

{0.186}^{\circ} C

2

{0.372}^{\circ} C

3

{1.86}^{\circ} C

4

{3.72}^{\circ} C

Explanation:

$Δ T_{f} = K_{f} \times m$
$Δ T_{f} = 1.86 \times \frac{0.02}{100} \times 1000$
$Δ T_{f} = 1.86 \times 0.02 \times 10$
$Δ T_{f} = {0.372}^{\circ} C$

NCERT-XII-18

4 RBTS PAPER

163830 A solution containing $8.6 g$ urea in one litre was found to be isotonic with $0.5 %$ (wt./vol) solution of an organic non volatile solute. The molecular weight of organic solute is:

1 348.9

2 34.89

3 3489

4 861.2

Explanation:

$π_{1} = π_{2}$
$C_{1} R T = C_{2} R T$
$C_{1} = C_{2}$
$\frac{8.6}{60} = \frac{0.5}{M \times 0.1}$
$M = 34.89$

NCERT-XII-21

4 RBTS PAPER

163825 Elevation in boiling point was ${0.52}^{\circ} C$ when $6 g$ of a compound $X$ was dissolved in $100 g$ of water. Molecular weight of $X$ is: $(Kb$ for water $= 0.52 K$ ${mol}^{- 1}$ )

1 120

2 60

3 100

4 342

Explanation:

$Δ T_{b} = K_{b} \times m$
$0.52 = 0.52 \times m$
$m = 1$
$\frac{6}{M_{X}} \times \frac{1000}{100} = 1$
$M_{x} = 60$

NCERT-XII-17

4 RBTS PAPER

163826 An aqueous solution containing $1 g$ of urea boils at ${100.25}^{\circ} C$ . The aqueous solution containing $3 g$ of glucose in the same volume will boil at -

1

{100.75}^{\circ} C

2

{100.5}^{\circ} C

3

100^{\circ} C

4

{100.25}^{\circ} C

Explanation:

For Urea solution
$Δ T_{b} = K_{b} (\frac{1}{60 \times W_{solvent}})$
$0.25 = K_{b} (\frac{1}{60 \times W_{solvent}})$

For glucose solution
$(T_{b} - 100) = K_{b} (\frac{1}{180 \times W_{solvent}})$

From $(2) \div (1)$
$\frac{T_{b} - 100}{0.25} = 1$
$T_{b} = 100 + 0.25$
$T_{b} = 100.25$

NCERT-XII-17

4 RBTS PAPER

163827 Pure benzene freezes at ${5.45}^{\circ} C$ at a certain place but a $0.374 m$ solution of tetrachloroethane in benzene freezes at ${3.55}^{\circ} C$ . The $K_{f}$ for benzene is-

1

5.08 K Kg {mol}^{- 1}

2

508 K Kg {mol}^{- 1}

3

0.508 K Kg {mol}^{- 1}

4

{50.8}^{\circ} C Kg {mol}^{- 1}

Explanation:

$Δ T_{f} = K_{f} \times m$
$T_{f}^{0} - T_{f} = K_{f} \times m$
$5.45 - 3.55 = K_{f} \times 0.374$
$K_{f} = \frac{1.9}{0.374} = 5.08 K Kg / mol$

NCERT-XII-18

4 RBTS PAPER

163828 Molal depression constant of water is $1.86 K Kg$ ${mol}^{- 1} .0 .02$ mole of urea dissolved in $100 g$ of water will produce a depression in freezing point of:

1

{0.186}^{\circ} C

2

{0.372}^{\circ} C

3

{1.86}^{\circ} C

4

{3.72}^{\circ} C

Explanation:

$Δ T_{f} = K_{f} \times m$
$Δ T_{f} = 1.86 \times \frac{0.02}{100} \times 1000$
$Δ T_{f} = 1.86 \times 0.02 \times 10$
$Δ T_{f} = {0.372}^{\circ} C$

NCERT-XII-18

4 RBTS PAPER

163830 A solution containing $8.6 g$ urea in one litre was found to be isotonic with $0.5 %$ (wt./vol) solution of an organic non volatile solute. The molecular weight of organic solute is:

1 348.9

2 34.89

3 3489

4 861.2

Explanation:

$π_{1} = π_{2}$
$C_{1} R T = C_{2} R T$
$C_{1} = C_{2}$
$\frac{8.6}{60} = \frac{0.5}{M \times 0.1}$
$M = 34.89$

NCERT-XII-21

4 RBTS PAPER

163825 Elevation in boiling point was ${0.52}^{\circ} C$ when $6 g$ of a compound $X$ was dissolved in $100 g$ of water. Molecular weight of $X$ is: $(Kb$ for water $= 0.52 K$ ${mol}^{- 1}$ )

1 120

2 60

3 100

4 342

Explanation:

$Δ T_{b} = K_{b} \times m$
$0.52 = 0.52 \times m$
$m = 1$
$\frac{6}{M_{X}} \times \frac{1000}{100} = 1$
$M_{x} = 60$

NCERT-XII-17

4 RBTS PAPER

163826 An aqueous solution containing $1 g$ of urea boils at ${100.25}^{\circ} C$ . The aqueous solution containing $3 g$ of glucose in the same volume will boil at -

1

{100.75}^{\circ} C

2

{100.5}^{\circ} C

3

100^{\circ} C

4

{100.25}^{\circ} C

Explanation:

For Urea solution
$Δ T_{b} = K_{b} (\frac{1}{60 \times W_{solvent}})$
$0.25 = K_{b} (\frac{1}{60 \times W_{solvent}})$

For glucose solution
$(T_{b} - 100) = K_{b} (\frac{1}{180 \times W_{solvent}})$

From $(2) \div (1)$
$\frac{T_{b} - 100}{0.25} = 1$
$T_{b} = 100 + 0.25$
$T_{b} = 100.25$

NCERT-XII-17

4 RBTS PAPER

163827 Pure benzene freezes at ${5.45}^{\circ} C$ at a certain place but a $0.374 m$ solution of tetrachloroethane in benzene freezes at ${3.55}^{\circ} C$ . The $K_{f}$ for benzene is-

1

5.08 K Kg {mol}^{- 1}

2

508 K Kg {mol}^{- 1}

3

0.508 K Kg {mol}^{- 1}

4

{50.8}^{\circ} C Kg {mol}^{- 1}

Explanation:

$Δ T_{f} = K_{f} \times m$
$T_{f}^{0} - T_{f} = K_{f} \times m$
$5.45 - 3.55 = K_{f} \times 0.374$
$K_{f} = \frac{1.9}{0.374} = 5.08 K Kg / mol$

NCERT-XII-18

4 RBTS PAPER

163828 Molal depression constant of water is $1.86 K Kg$ ${mol}^{- 1} .0 .02$ mole of urea dissolved in $100 g$ of water will produce a depression in freezing point of:

1

{0.186}^{\circ} C

2

{0.372}^{\circ} C

3

{1.86}^{\circ} C

4

{3.72}^{\circ} C

Explanation:

$Δ T_{f} = K_{f} \times m$
$Δ T_{f} = 1.86 \times \frac{0.02}{100} \times 1000$
$Δ T_{f} = 1.86 \times 0.02 \times 10$
$Δ T_{f} = {0.372}^{\circ} C$

NCERT-XII-18

4 RBTS PAPER

163830 A solution containing $8.6 g$ urea in one litre was found to be isotonic with $0.5 %$ (wt./vol) solution of an organic non volatile solute. The molecular weight of organic solute is:

1 348.9

2 34.89

3 3489

4 861.2

Explanation:

$π_{1} = π_{2}$
$C_{1} R T = C_{2} R T$
$C_{1} = C_{2}$
$\frac{8.6}{60} = \frac{0.5}{M \times 0.1}$
$M = 34.89$

NCERT-XII-21

4 RBTS PAPER

163825 Elevation in boiling point was ${0.52}^{\circ} C$ when $6 g$ of a compound $X$ was dissolved in $100 g$ of water. Molecular weight of $X$ is: $(Kb$ for water $= 0.52 K$ ${mol}^{- 1}$ )

1 120

2 60

3 100

4 342

Explanation:

$Δ T_{b} = K_{b} \times m$
$0.52 = 0.52 \times m$
$m = 1$
$\frac{6}{M_{X}} \times \frac{1000}{100} = 1$
$M_{x} = 60$

NCERT-XII-17

4 RBTS PAPER

163826 An aqueous solution containing $1 g$ of urea boils at ${100.25}^{\circ} C$ . The aqueous solution containing $3 g$ of glucose in the same volume will boil at -

1

{100.75}^{\circ} C

2

{100.5}^{\circ} C

3

100^{\circ} C

4

{100.25}^{\circ} C

Explanation:

For Urea solution
$Δ T_{b} = K_{b} (\frac{1}{60 \times W_{solvent}})$
$0.25 = K_{b} (\frac{1}{60 \times W_{solvent}})$

For glucose solution
$(T_{b} - 100) = K_{b} (\frac{1}{180 \times W_{solvent}})$

From $(2) \div (1)$
$\frac{T_{b} - 100}{0.25} = 1$
$T_{b} = 100 + 0.25$
$T_{b} = 100.25$

NCERT-XII-17

4 RBTS PAPER

163827 Pure benzene freezes at ${5.45}^{\circ} C$ at a certain place but a $0.374 m$ solution of tetrachloroethane in benzene freezes at ${3.55}^{\circ} C$ . The $K_{f}$ for benzene is-

1

5.08 K Kg {mol}^{- 1}

2

508 K Kg {mol}^{- 1}

3

0.508 K Kg {mol}^{- 1}

4

{50.8}^{\circ} C Kg {mol}^{- 1}

Explanation:

$Δ T_{f} = K_{f} \times m$
$T_{f}^{0} - T_{f} = K_{f} \times m$
$5.45 - 3.55 = K_{f} \times 0.374$
$K_{f} = \frac{1.9}{0.374} = 5.08 K Kg / mol$

NCERT-XII-18

4 RBTS PAPER

163828 Molal depression constant of water is $1.86 K Kg$ ${mol}^{- 1} .0 .02$ mole of urea dissolved in $100 g$ of water will produce a depression in freezing point of:

1

{0.186}^{\circ} C

2

{0.372}^{\circ} C

3

{1.86}^{\circ} C

4

{3.72}^{\circ} C

Explanation:

$Δ T_{f} = K_{f} \times m$
$Δ T_{f} = 1.86 \times \frac{0.02}{100} \times 1000$
$Δ T_{f} = 1.86 \times 0.02 \times 10$
$Δ T_{f} = {0.372}^{\circ} C$

NCERT-XII-18

4 RBTS PAPER

163830 A solution containing $8.6 g$ urea in one litre was found to be isotonic with $0.5 %$ (wt./vol) solution of an organic non volatile solute. The molecular weight of organic solute is:

1 348.9

2 34.89

3 3489

4 861.2

Explanation:

$π_{1} = π_{2}$
$C_{1} R T = C_{2} R T$
$C_{1} = C_{2}$
$\frac{8.6}{60} = \frac{0.5}{M \times 0.1}$
$M = 34.89$

NCERT-XII-21