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CHXII02:SOLUTIONS

319224 $4.0 g$ of $N a O H$ is dissolved in $100 m L$ solution. The normality of the solution is

1

0.1 N

2

0.5 N

3

4.0 N

4

1.0 N

Explanation:

Given, mass of solute $= 4.0 g$
$Normality = \frac{gram equivalent of solute}{volume of solution (in L)}$
Molecular mass of $NaOH = 23 + 16 + 1$
$= 40$
Volume of solution
$= 100 m L = 100 / 1000 L$
$= 0.1 L$
$∴ N o r m a l i t y = \frac{4 / 40}{0.1} = 1.0 N$

CHXII02:SOLUTIONS

319225 10 g of NaCl is dissolved in $1 0^{6} g$ of the solution. Its concentration is:

1 100 ppm

2 0.1 ppm

3 1 ppm

4 10 ppm

Explanation:

$p p m = \frac{M a s s o f s o l u t e}{M a s s o f s o l u t i o n} \times 1 0^{6}$
$∴ p p m = \frac{10}{1 0^{6}} \times 1 0^{6} = 10 p p m$

CHXII02:SOLUTIONS

319226 Assertion :
One molar aqueous solution has always higher concentration than one molal.
Reason :
The molality of a solution depends upon the density of the solution whereas molarity does not.

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.

Explanation:

Molarity is one mole solute dissolved in 1L of solution. Whereas, molality is one mole solute disssolved in $1 Kg$ solvent. Since volume of solution includes volume of both solute and solvent, the solute is present in lesser quantity in molar solution than that present in molal solution. Molality depends upon the density of the solution whereas molarity does not. So option (2) is correct.

CHXII02:SOLUTIONS

319227 A 5.2 molal aqueous solution of $C H_{3} O H$ is supplied. What is the mole fraction of methyl alcohol in the solution?

1 0.086

2

0.190

3

0.050

4

0.100

Explanation:

We know, $m = \frac{x_{B} \times 1000}{(1 - x_{B}) m_{A}}$
Given: $m = 5 .2 a n d m_{A} = 18$
$∴ 5 .2 = \frac{x_{B} \times 1000}{(1 - x_{B}) \times 18}$
$o r x_{B} = 0 .086$

CHXII02:SOLUTIONS

319224 $4.0 g$ of $N a O H$ is dissolved in $100 m L$ solution. The normality of the solution is

1

0.1 N

2

0.5 N

3

4.0 N

4

1.0 N

Explanation:

Given, mass of solute $= 4.0 g$
$Normality = \frac{gram equivalent of solute}{volume of solution (in L)}$
Molecular mass of $NaOH = 23 + 16 + 1$
$= 40$
Volume of solution
$= 100 m L = 100 / 1000 L$
$= 0.1 L$
$∴ N o r m a l i t y = \frac{4 / 40}{0.1} = 1.0 N$

CHXII02:SOLUTIONS

319225 10 g of NaCl is dissolved in $1 0^{6} g$ of the solution. Its concentration is:

1 100 ppm

2 0.1 ppm

3 1 ppm

4 10 ppm

Explanation:

$p p m = \frac{M a s s o f s o l u t e}{M a s s o f s o l u t i o n} \times 1 0^{6}$
$∴ p p m = \frac{10}{1 0^{6}} \times 1 0^{6} = 10 p p m$

CHXII02:SOLUTIONS

319226 Assertion :
One molar aqueous solution has always higher concentration than one molal.
Reason :
The molality of a solution depends upon the density of the solution whereas molarity does not.

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.

Explanation:

Molarity is one mole solute dissolved in 1L of solution. Whereas, molality is one mole solute disssolved in $1 Kg$ solvent. Since volume of solution includes volume of both solute and solvent, the solute is present in lesser quantity in molar solution than that present in molal solution. Molality depends upon the density of the solution whereas molarity does not. So option (2) is correct.

CHXII02:SOLUTIONS

319227 A 5.2 molal aqueous solution of $C H_{3} O H$ is supplied. What is the mole fraction of methyl alcohol in the solution?

1 0.086

2

0.190

3

0.050

4

0.100

Explanation:

We know, $m = \frac{x_{B} \times 1000}{(1 - x_{B}) m_{A}}$
Given: $m = 5 .2 a n d m_{A} = 18$
$∴ 5 .2 = \frac{x_{B} \times 1000}{(1 - x_{B}) \times 18}$
$o r x_{B} = 0 .086$

CHXII02:SOLUTIONS

319224 $4.0 g$ of $N a O H$ is dissolved in $100 m L$ solution. The normality of the solution is

1

0.1 N

2

0.5 N

3

4.0 N

4

1.0 N

Explanation:

Given, mass of solute $= 4.0 g$
$Normality = \frac{gram equivalent of solute}{volume of solution (in L)}$
Molecular mass of $NaOH = 23 + 16 + 1$
$= 40$
Volume of solution
$= 100 m L = 100 / 1000 L$
$= 0.1 L$
$∴ N o r m a l i t y = \frac{4 / 40}{0.1} = 1.0 N$

CHXII02:SOLUTIONS

319225 10 g of NaCl is dissolved in $1 0^{6} g$ of the solution. Its concentration is:

1 100 ppm

2 0.1 ppm

3 1 ppm

4 10 ppm

Explanation:

$p p m = \frac{M a s s o f s o l u t e}{M a s s o f s o l u t i o n} \times 1 0^{6}$
$∴ p p m = \frac{10}{1 0^{6}} \times 1 0^{6} = 10 p p m$

CHXII02:SOLUTIONS

319226 Assertion :
One molar aqueous solution has always higher concentration than one molal.
Reason :
The molality of a solution depends upon the density of the solution whereas molarity does not.

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.

Explanation:

Molarity is one mole solute dissolved in 1L of solution. Whereas, molality is one mole solute disssolved in $1 Kg$ solvent. Since volume of solution includes volume of both solute and solvent, the solute is present in lesser quantity in molar solution than that present in molal solution. Molality depends upon the density of the solution whereas molarity does not. So option (2) is correct.

CHXII02:SOLUTIONS

319227 A 5.2 molal aqueous solution of $C H_{3} O H$ is supplied. What is the mole fraction of methyl alcohol in the solution?

1 0.086

2

0.190

3

0.050

4

0.100

Explanation:

We know, $m = \frac{x_{B} \times 1000}{(1 - x_{B}) m_{A}}$
Given: $m = 5 .2 a n d m_{A} = 18$
$∴ 5 .2 = \frac{x_{B} \times 1000}{(1 - x_{B}) \times 18}$
$o r x_{B} = 0 .086$

CHXII02:SOLUTIONS

319224 $4.0 g$ of $N a O H$ is dissolved in $100 m L$ solution. The normality of the solution is

1

0.1 N

2

0.5 N

3

4.0 N

4

1.0 N

Explanation:

Given, mass of solute $= 4.0 g$
$Normality = \frac{gram equivalent of solute}{volume of solution (in L)}$
Molecular mass of $NaOH = 23 + 16 + 1$
$= 40$
Volume of solution
$= 100 m L = 100 / 1000 L$
$= 0.1 L$
$∴ N o r m a l i t y = \frac{4 / 40}{0.1} = 1.0 N$

CHXII02:SOLUTIONS

319225 10 g of NaCl is dissolved in $1 0^{6} g$ of the solution. Its concentration is:

1 100 ppm

2 0.1 ppm

3 1 ppm

4 10 ppm

Explanation:

$p p m = \frac{M a s s o f s o l u t e}{M a s s o f s o l u t i o n} \times 1 0^{6}$
$∴ p p m = \frac{10}{1 0^{6}} \times 1 0^{6} = 10 p p m$

CHXII02:SOLUTIONS

319226 Assertion :
One molar aqueous solution has always higher concentration than one molal.
Reason :
The molality of a solution depends upon the density of the solution whereas molarity does not.

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.

Explanation:

Molarity is one mole solute dissolved in 1L of solution. Whereas, molality is one mole solute disssolved in $1 Kg$ solvent. Since volume of solution includes volume of both solute and solvent, the solute is present in lesser quantity in molar solution than that present in molal solution. Molality depends upon the density of the solution whereas molarity does not. So option (2) is correct.

CHXII02:SOLUTIONS

319227 A 5.2 molal aqueous solution of $C H_{3} O H$ is supplied. What is the mole fraction of methyl alcohol in the solution?

1 0.086

2

0.190

3

0.050

4

0.100

Explanation:

We know, $m = \frac{x_{B} \times 1000}{(1 - x_{B}) m_{A}}$
Given: $m = 5 .2 a n d m_{A} = 18$
$∴ 5 .2 = \frac{x_{B} \times 1000}{(1 - x_{B}) \times 18}$
$o r x_{B} = 0 .086$