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CHXII02:SOLUTIONS

319228 A solution of $A l_{2} {(S O_{4})}_{3} {d = 1 .253 g m / m l}$ contains 22% salt by weight. The molarity, normality, and molality of the solution is, respectively

1 0.825 M, 48.3 N, 0.805 m

2 0.805 M, 4.83 N, 0.825 m

3 4.83 M, 4.83 N, 4.83 m

4 None

Explanation:

$M o l a r i t y = \frac{% \times 10 \times d}{M W_{2}} = \frac{22 \times 10 \times 1 .253}{342}$
$= 0 .805 M$
$N o r m a l i t y = \frac{% \times 10 \times d}{E W_{2}} = \frac{22 \times 10 \times 1 .253}{342 / 6} = 4 .83 N$
$M o l a l i t y = \frac{22 \times 1000}{342 (100 - 22)} = 0 .825 m$

CHXII02:SOLUTIONS

319229 A solution of sucrose in water is labelled as 20% w/w. What should be the mole fraction of sucrose in the solution?

1 0.013

2 0.987

3 0.13

4 9.87

Explanation:

$X_{2} = \frac{n_{2}}{n_{1} + n_{2}}$
$= \frac{(20 / 180)}{(20 / 180) + (80 / 18)} = 0 .013$

CHXII02:SOLUTIONS

319230 Assertion :
When a concentrated solution is diluted by adding more water, the number of moles of solute in the solution remains unchanged.
Reason :
Number of moles of a solute is equal to the product of molarity and volume of solution in litres.

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.

CHXII02:SOLUTIONS

319231 Molarity of $0.1 N$ oxalic acid is

1

0.05 M

2

0.01 M

3

0.2 M

4

0.3 M

Explanation:

Oxalic acid is a dibasic acid.
$N = M \times$ basicity
$M = \frac{N}{Basicity} = \frac{0.1}{2} = 0.05 M$

CHXII02:SOLUTIONS

319228 A solution of $A l_{2} {(S O_{4})}_{3} {d = 1 .253 g m / m l}$ contains 22% salt by weight. The molarity, normality, and molality of the solution is, respectively

1 0.825 M, 48.3 N, 0.805 m

2 0.805 M, 4.83 N, 0.825 m

3 4.83 M, 4.83 N, 4.83 m

4 None

Explanation:

$M o l a r i t y = \frac{% \times 10 \times d}{M W_{2}} = \frac{22 \times 10 \times 1 .253}{342}$
$= 0 .805 M$
$N o r m a l i t y = \frac{% \times 10 \times d}{E W_{2}} = \frac{22 \times 10 \times 1 .253}{342 / 6} = 4 .83 N$
$M o l a l i t y = \frac{22 \times 1000}{342 (100 - 22)} = 0 .825 m$

CHXII02:SOLUTIONS

319229 A solution of sucrose in water is labelled as 20% w/w. What should be the mole fraction of sucrose in the solution?

1 0.013

2 0.987

3 0.13

4 9.87

Explanation:

$X_{2} = \frac{n_{2}}{n_{1} + n_{2}}$
$= \frac{(20 / 180)}{(20 / 180) + (80 / 18)} = 0 .013$

CHXII02:SOLUTIONS

319230 Assertion :
When a concentrated solution is diluted by adding more water, the number of moles of solute in the solution remains unchanged.
Reason :
Number of moles of a solute is equal to the product of molarity and volume of solution in litres.

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.

CHXII02:SOLUTIONS

319231 Molarity of $0.1 N$ oxalic acid is

1

0.05 M

2

0.01 M

3

0.2 M

4

0.3 M

Explanation:

Oxalic acid is a dibasic acid.
$N = M \times$ basicity
$M = \frac{N}{Basicity} = \frac{0.1}{2} = 0.05 M$

CHXII02:SOLUTIONS

319228 A solution of $A l_{2} {(S O_{4})}_{3} {d = 1 .253 g m / m l}$ contains 22% salt by weight. The molarity, normality, and molality of the solution is, respectively

1 0.825 M, 48.3 N, 0.805 m

2 0.805 M, 4.83 N, 0.825 m

3 4.83 M, 4.83 N, 4.83 m

4 None

Explanation:

$M o l a r i t y = \frac{% \times 10 \times d}{M W_{2}} = \frac{22 \times 10 \times 1 .253}{342}$
$= 0 .805 M$
$N o r m a l i t y = \frac{% \times 10 \times d}{E W_{2}} = \frac{22 \times 10 \times 1 .253}{342 / 6} = 4 .83 N$
$M o l a l i t y = \frac{22 \times 1000}{342 (100 - 22)} = 0 .825 m$

CHXII02:SOLUTIONS

319229 A solution of sucrose in water is labelled as 20% w/w. What should be the mole fraction of sucrose in the solution?

1 0.013

2 0.987

3 0.13

4 9.87

Explanation:

$X_{2} = \frac{n_{2}}{n_{1} + n_{2}}$
$= \frac{(20 / 180)}{(20 / 180) + (80 / 18)} = 0 .013$

CHXII02:SOLUTIONS

319230 Assertion :
When a concentrated solution is diluted by adding more water, the number of moles of solute in the solution remains unchanged.
Reason :
Number of moles of a solute is equal to the product of molarity and volume of solution in litres.

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.

CHXII02:SOLUTIONS

319231 Molarity of $0.1 N$ oxalic acid is

1

0.05 M

2

0.01 M

3

0.2 M

4

0.3 M

Explanation:

Oxalic acid is a dibasic acid.
$N = M \times$ basicity
$M = \frac{N}{Basicity} = \frac{0.1}{2} = 0.05 M$

CHXII02:SOLUTIONS

319228 A solution of $A l_{2} {(S O_{4})}_{3} {d = 1 .253 g m / m l}$ contains 22% salt by weight. The molarity, normality, and molality of the solution is, respectively

1 0.825 M, 48.3 N, 0.805 m

2 0.805 M, 4.83 N, 0.825 m

3 4.83 M, 4.83 N, 4.83 m

4 None

Explanation:

$M o l a r i t y = \frac{% \times 10 \times d}{M W_{2}} = \frac{22 \times 10 \times 1 .253}{342}$
$= 0 .805 M$
$N o r m a l i t y = \frac{% \times 10 \times d}{E W_{2}} = \frac{22 \times 10 \times 1 .253}{342 / 6} = 4 .83 N$
$M o l a l i t y = \frac{22 \times 1000}{342 (100 - 22)} = 0 .825 m$

CHXII02:SOLUTIONS

319229 A solution of sucrose in water is labelled as 20% w/w. What should be the mole fraction of sucrose in the solution?

1 0.013

2 0.987

3 0.13

4 9.87

Explanation:

$X_{2} = \frac{n_{2}}{n_{1} + n_{2}}$
$= \frac{(20 / 180)}{(20 / 180) + (80 / 18)} = 0 .013$

CHXII02:SOLUTIONS

319230 Assertion :
When a concentrated solution is diluted by adding more water, the number of moles of solute in the solution remains unchanged.
Reason :
Number of moles of a solute is equal to the product of molarity and volume of solution in litres.

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.

CHXII02:SOLUTIONS

319231 Molarity of $0.1 N$ oxalic acid is

1

0.05 M

2

0.01 M

3

0.2 M

4

0.3 M

Explanation:

Oxalic acid is a dibasic acid.
$N = M \times$ basicity
$M = \frac{N}{Basicity} = \frac{0.1}{2} = 0.05 M$

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