QBManager

4 RBTS PAPER

163845 During freezing of an aqueous solution of sodium chloride, there exists an equilibrium between :

1 Solid solute and solid solvent

2 Solid solvent and liquid solvent

3 Liquid solute and liquid solvent

4 Liquid solute and solid solvent

NCERT-XII-18

4 RBTS PAPER

163829 The osmotic pressure of a solution increases if:

1 Temperature is lowered

2 Volume is increased

3 Number of solute molecules is increased

4 None

NCERT-XII-15

4 RBTS PAPER

163846 The von't Hoff factor for $0.1 M {Cs}_{2} {SO}_{4}$ solution is 2.74. The degree of dissociation of salt will be :

1

34.8 %

2

87.0 %

3

91.3 %

4

98.4 %

Explanation:

$i = 1 - α + 2 α + α$
$2.74 = 1 + 2 α$
$α = 0.87$
$% α = 87 %$

NCERT-XII-24

4 RBTS PAPER

163815 For preparing $0.1 M$ solution of $H_{2} {SO}_{4}$ in one litre, we need $H_{2} {SO}_{4}$ :

1

0.98 g

2

4.9 g

3

49.0 g

4

9.8 g

Explanation:

$moles = M \times V$
$moles = 0.1 \times 1 = 0.1$
$\frac{W}{98} = 0.1$
$W = 9.8 g$

NCERT-XII-04

4 RBTS PAPER

163816 An $X$ molal solution of a compound in benzene has mole fraction of solute equal to $0.2$ . The value of $X$ is:

1 14

2 3.2

3 1.4

4 2

Explanation:

Let moles of solute $= n$
$&$ moles of solvent $= N$
$X_{solute} = \frac{n}{n + N} = 0.2$
$∴ \frac{n}{N} = \frac{1}{4}$
$molality = \frac{moles of solute (n)}{wt. of solvent (C_{2} H_{6}) (kg)}$
$= \frac{1 mol}{(4 \times 78 gm)} \times 1000 = 3.2$

NCERT-XII-04

4 RBTS PAPER

163845 During freezing of an aqueous solution of sodium chloride, there exists an equilibrium between :

1 Solid solute and solid solvent

2 Solid solvent and liquid solvent

3 Liquid solute and liquid solvent

4 Liquid solute and solid solvent

NCERT-XII-18

4 RBTS PAPER

163829 The osmotic pressure of a solution increases if:

1 Temperature is lowered

2 Volume is increased

3 Number of solute molecules is increased

4 None

NCERT-XII-15

4 RBTS PAPER

163846 The von't Hoff factor for $0.1 M {Cs}_{2} {SO}_{4}$ solution is 2.74. The degree of dissociation of salt will be :

1

34.8 %

2

87.0 %

3

91.3 %

4

98.4 %

Explanation:

$i = 1 - α + 2 α + α$
$2.74 = 1 + 2 α$
$α = 0.87$
$% α = 87 %$

NCERT-XII-24

4 RBTS PAPER

163815 For preparing $0.1 M$ solution of $H_{2} {SO}_{4}$ in one litre, we need $H_{2} {SO}_{4}$ :

1

0.98 g

2

4.9 g

3

49.0 g

4

9.8 g

Explanation:

$moles = M \times V$
$moles = 0.1 \times 1 = 0.1$
$\frac{W}{98} = 0.1$
$W = 9.8 g$

NCERT-XII-04

4 RBTS PAPER

163816 An $X$ molal solution of a compound in benzene has mole fraction of solute equal to $0.2$ . The value of $X$ is:

1 14

2 3.2

3 1.4

4 2

Explanation:

Let moles of solute $= n$
$&$ moles of solvent $= N$
$X_{solute} = \frac{n}{n + N} = 0.2$
$∴ \frac{n}{N} = \frac{1}{4}$
$molality = \frac{moles of solute (n)}{wt. of solvent (C_{2} H_{6}) (kg)}$
$= \frac{1 mol}{(4 \times 78 gm)} \times 1000 = 3.2$

NCERT-XII-04

4 RBTS PAPER

163845 During freezing of an aqueous solution of sodium chloride, there exists an equilibrium between :

1 Solid solute and solid solvent

2 Solid solvent and liquid solvent

3 Liquid solute and liquid solvent

4 Liquid solute and solid solvent

NCERT-XII-18

4 RBTS PAPER

163829 The osmotic pressure of a solution increases if:

1 Temperature is lowered

2 Volume is increased

3 Number of solute molecules is increased

4 None

NCERT-XII-15

4 RBTS PAPER

163846 The von't Hoff factor for $0.1 M {Cs}_{2} {SO}_{4}$ solution is 2.74. The degree of dissociation of salt will be :

1

34.8 %

2

87.0 %

3

91.3 %

4

98.4 %

Explanation:

$i = 1 - α + 2 α + α$
$2.74 = 1 + 2 α$
$α = 0.87$
$% α = 87 %$

NCERT-XII-24

4 RBTS PAPER

163815 For preparing $0.1 M$ solution of $H_{2} {SO}_{4}$ in one litre, we need $H_{2} {SO}_{4}$ :

1

0.98 g

2

4.9 g

3

49.0 g

4

9.8 g

Explanation:

$moles = M \times V$
$moles = 0.1 \times 1 = 0.1$
$\frac{W}{98} = 0.1$
$W = 9.8 g$

NCERT-XII-04

4 RBTS PAPER

163816 An $X$ molal solution of a compound in benzene has mole fraction of solute equal to $0.2$ . The value of $X$ is:

1 14

2 3.2

3 1.4

4 2

Explanation:

Let moles of solute $= n$
$&$ moles of solvent $= N$
$X_{solute} = \frac{n}{n + N} = 0.2$
$∴ \frac{n}{N} = \frac{1}{4}$
$molality = \frac{moles of solute (n)}{wt. of solvent (C_{2} H_{6}) (kg)}$
$= \frac{1 mol}{(4 \times 78 gm)} \times 1000 = 3.2$

NCERT-XII-04

4 RBTS PAPER

163845 During freezing of an aqueous solution of sodium chloride, there exists an equilibrium between :

1 Solid solute and solid solvent

2 Solid solvent and liquid solvent

3 Liquid solute and liquid solvent

4 Liquid solute and solid solvent

NCERT-XII-18

4 RBTS PAPER

163829 The osmotic pressure of a solution increases if:

1 Temperature is lowered

2 Volume is increased

3 Number of solute molecules is increased

4 None

NCERT-XII-15

4 RBTS PAPER

163846 The von't Hoff factor for $0.1 M {Cs}_{2} {SO}_{4}$ solution is 2.74. The degree of dissociation of salt will be :

1

34.8 %

2

87.0 %

3

91.3 %

4

98.4 %

Explanation:

$i = 1 - α + 2 α + α$
$2.74 = 1 + 2 α$
$α = 0.87$
$% α = 87 %$

NCERT-XII-24

4 RBTS PAPER

163815 For preparing $0.1 M$ solution of $H_{2} {SO}_{4}$ in one litre, we need $H_{2} {SO}_{4}$ :

1

0.98 g

2

4.9 g

3

49.0 g

4

9.8 g

Explanation:

$moles = M \times V$
$moles = 0.1 \times 1 = 0.1$
$\frac{W}{98} = 0.1$
$W = 9.8 g$

NCERT-XII-04

4 RBTS PAPER

163816 An $X$ molal solution of a compound in benzene has mole fraction of solute equal to $0.2$ . The value of $X$ is:

1 14

2 3.2

3 1.4

4 2

Explanation:

Let moles of solute $= n$
$&$ moles of solvent $= N$
$X_{solute} = \frac{n}{n + N} = 0.2$
$∴ \frac{n}{N} = \frac{1}{4}$
$molality = \frac{moles of solute (n)}{wt. of solvent (C_{2} H_{6}) (kg)}$
$= \frac{1 mol}{(4 \times 78 gm)} \times 1000 = 3.2$

NCERT-XII-04

4 RBTS PAPER

163845 During freezing of an aqueous solution of sodium chloride, there exists an equilibrium between :

1 Solid solute and solid solvent

2 Solid solvent and liquid solvent

3 Liquid solute and liquid solvent

4 Liquid solute and solid solvent

NCERT-XII-18

4 RBTS PAPER

163829 The osmotic pressure of a solution increases if:

1 Temperature is lowered

2 Volume is increased

3 Number of solute molecules is increased

4 None

NCERT-XII-15

4 RBTS PAPER

163846 The von't Hoff factor for $0.1 M {Cs}_{2} {SO}_{4}$ solution is 2.74. The degree of dissociation of salt will be :

1

34.8 %

2

87.0 %

3

91.3 %

4

98.4 %

Explanation:

$i = 1 - α + 2 α + α$
$2.74 = 1 + 2 α$
$α = 0.87$
$% α = 87 %$

NCERT-XII-24

4 RBTS PAPER

163815 For preparing $0.1 M$ solution of $H_{2} {SO}_{4}$ in one litre, we need $H_{2} {SO}_{4}$ :

1

0.98 g

2

4.9 g

3

49.0 g

4

9.8 g

Explanation:

$moles = M \times V$
$moles = 0.1 \times 1 = 0.1$
$\frac{W}{98} = 0.1$
$W = 9.8 g$

NCERT-XII-04

4 RBTS PAPER

163816 An $X$ molal solution of a compound in benzene has mole fraction of solute equal to $0.2$ . The value of $X$ is:

1 14

2 3.2

3 1.4

4 2

Explanation:

Let moles of solute $= n$
$&$ moles of solvent $= N$
$X_{solute} = \frac{n}{n + N} = 0.2$
$∴ \frac{n}{N} = \frac{1}{4}$
$molality = \frac{moles of solute (n)}{wt. of solvent (C_{2} H_{6}) (kg)}$
$= \frac{1 mol}{(4 \times 78 gm)} \times 1000 = 3.2$

NCERT-XII-04