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CHXII02:SOLUTIONS

319207 The molarity of a solution obtained by mixing of 750 mL of 0.5 M HCl with 250 mL of 2 M HCl will be

1 0.875 M

2 1 M

3 1.25 M

4 2.5 M

Explanation:

$M_{M i x t u r e} = \frac{M_{1} V_{1} + M_{2} V_{2}}{V_{1} + V_{2}}$
$= \frac{0 .5 \times 750 + 2 \times 250}{750 + 250} = 0 .875 M$

CHXII02:SOLUTIONS

319208 A student has 100 mL of 0.1 M KCl solution. To make it 0.2 M, he has to

1 evaporate 50 mL of the solution

2 add 0.01 mole of KCl

3 both (1) and (2) can be used

4 neither (1) nor (2) can be used

Explanation:

100 mL of 0.1 M KCl contains 0.01 mole of KCl
On evaporating 50 mL of solution, now conc.
$= \frac{0 .01}{50} \times 1000 = 0 .2 M$
On adding 0.01 mole of KCl, now KCl present = 0.02 mole
$∴$ Concentration $= \frac{0 .02}{100} \times 1000 = 0 .2 M$
Thus, both (1) and (2) can be used.

CHXII02:SOLUTIONS

319209 The molarity of urea (molar mass $60 g m o l^{- 1}$ ) solution by dissolving 15 g of urea in $500 c m^{3}$ of water is

1

2 m o l d m^{- 3}

2

0 .5 m o l d m^{- 3}

3

0 .125 m o l d m^{- 3}

4

0 .0005 m o l d m^{- 3}

Explanation:

Given, mass of urea = 15 g
Molar mass of urea $= 60 g m o l^{- 1}$
$V o l u m e = 500 c m^{3} = 500 m L = 0 .5 L$
$∵$ $1 c m^{3} = 1 m L = 1000 L$
$∴ M o l a r i t y$
$= \frac{M a s s o f s o l u t e i n g r a m s (w)}{M o l a r m a s s o f s o l u t e (M) \times V o l u m e o f s o l u t i o n V (L))}$
$M o l a r i t y o f u r e a = \frac{15 g}{60 g m o l^{- 1} \times 0 .5 L}$
$\Rightarrow 0 .5 m o l / L = 0 .5 m o l / d m^{3}$
$(1 L = 1 d m^{3})$

MHTCET - 2018

CHXII02:SOLUTIONS

319210 Three statements are given about mole fraction
i. Mole fraction of a solute + mole fraction of solvent = 1
ii. Equal weights of Helium and methane are present in a gaseous mixture. The mole fraction of He is 4/5
iii. The mole fraction of $40 %$ by mass of NaOH solution is 0.5

1 i and ii are correct

2 ii and iii are correct

3 i and iii are correct

4 all are correct

Explanation:

(i) Sum of the mole fraction of all the components in solution is 1.
(ii) Let us assume weight of $H E a n d C H_{4}$ as 1 g Mole fraction of $H E = \frac{\frac{1}{4}}{\frac{1}{4} + \frac{1}{6}} = \frac{4}{5}$ .
(iii) 40% by mass of NaOH means 40 g of NaOH present in 100 g of solution.
Mole fraction of $N a O H = \frac{\frac{40}{40}}{(\frac{40}{40}) + (\frac{60}{18})} = 0 .23$

CHXII02:SOLUTIONS

319207 The molarity of a solution obtained by mixing of 750 mL of 0.5 M HCl with 250 mL of 2 M HCl will be

1 0.875 M

2 1 M

3 1.25 M

4 2.5 M

Explanation:

$M_{M i x t u r e} = \frac{M_{1} V_{1} + M_{2} V_{2}}{V_{1} + V_{2}}$
$= \frac{0 .5 \times 750 + 2 \times 250}{750 + 250} = 0 .875 M$

CHXII02:SOLUTIONS

319208 A student has 100 mL of 0.1 M KCl solution. To make it 0.2 M, he has to

1 evaporate 50 mL of the solution

2 add 0.01 mole of KCl

3 both (1) and (2) can be used

4 neither (1) nor (2) can be used

Explanation:

100 mL of 0.1 M KCl contains 0.01 mole of KCl
On evaporating 50 mL of solution, now conc.
$= \frac{0 .01}{50} \times 1000 = 0 .2 M$
On adding 0.01 mole of KCl, now KCl present = 0.02 mole
$∴$ Concentration $= \frac{0 .02}{100} \times 1000 = 0 .2 M$
Thus, both (1) and (2) can be used.

CHXII02:SOLUTIONS

319209 The molarity of urea (molar mass $60 g m o l^{- 1}$ ) solution by dissolving 15 g of urea in $500 c m^{3}$ of water is

1

2 m o l d m^{- 3}

2

0 .5 m o l d m^{- 3}

3

0 .125 m o l d m^{- 3}

4

0 .0005 m o l d m^{- 3}

Explanation:

Given, mass of urea = 15 g
Molar mass of urea $= 60 g m o l^{- 1}$
$V o l u m e = 500 c m^{3} = 500 m L = 0 .5 L$
$∵$ $1 c m^{3} = 1 m L = 1000 L$
$∴ M o l a r i t y$
$= \frac{M a s s o f s o l u t e i n g r a m s (w)}{M o l a r m a s s o f s o l u t e (M) \times V o l u m e o f s o l u t i o n V (L))}$
$M o l a r i t y o f u r e a = \frac{15 g}{60 g m o l^{- 1} \times 0 .5 L}$
$\Rightarrow 0 .5 m o l / L = 0 .5 m o l / d m^{3}$
$(1 L = 1 d m^{3})$

MHTCET - 2018

CHXII02:SOLUTIONS

319210 Three statements are given about mole fraction
i. Mole fraction of a solute + mole fraction of solvent = 1
ii. Equal weights of Helium and methane are present in a gaseous mixture. The mole fraction of He is 4/5
iii. The mole fraction of $40 %$ by mass of NaOH solution is 0.5

1 i and ii are correct

2 ii and iii are correct

3 i and iii are correct

4 all are correct

Explanation:

(i) Sum of the mole fraction of all the components in solution is 1.
(ii) Let us assume weight of $H E a n d C H_{4}$ as 1 g Mole fraction of $H E = \frac{\frac{1}{4}}{\frac{1}{4} + \frac{1}{6}} = \frac{4}{5}$ .
(iii) 40% by mass of NaOH means 40 g of NaOH present in 100 g of solution.
Mole fraction of $N a O H = \frac{\frac{40}{40}}{(\frac{40}{40}) + (\frac{60}{18})} = 0 .23$

CHXII02:SOLUTIONS

319207 The molarity of a solution obtained by mixing of 750 mL of 0.5 M HCl with 250 mL of 2 M HCl will be

1 0.875 M

2 1 M

3 1.25 M

4 2.5 M

Explanation:

$M_{M i x t u r e} = \frac{M_{1} V_{1} + M_{2} V_{2}}{V_{1} + V_{2}}$
$= \frac{0 .5 \times 750 + 2 \times 250}{750 + 250} = 0 .875 M$

CHXII02:SOLUTIONS

319208 A student has 100 mL of 0.1 M KCl solution. To make it 0.2 M, he has to

1 evaporate 50 mL of the solution

2 add 0.01 mole of KCl

3 both (1) and (2) can be used

4 neither (1) nor (2) can be used

Explanation:

100 mL of 0.1 M KCl contains 0.01 mole of KCl
On evaporating 50 mL of solution, now conc.
$= \frac{0 .01}{50} \times 1000 = 0 .2 M$
On adding 0.01 mole of KCl, now KCl present = 0.02 mole
$∴$ Concentration $= \frac{0 .02}{100} \times 1000 = 0 .2 M$
Thus, both (1) and (2) can be used.

CHXII02:SOLUTIONS

319209 The molarity of urea (molar mass $60 g m o l^{- 1}$ ) solution by dissolving 15 g of urea in $500 c m^{3}$ of water is

1

2 m o l d m^{- 3}

2

0 .5 m o l d m^{- 3}

3

0 .125 m o l d m^{- 3}

4

0 .0005 m o l d m^{- 3}

Explanation:

Given, mass of urea = 15 g
Molar mass of urea $= 60 g m o l^{- 1}$
$V o l u m e = 500 c m^{3} = 500 m L = 0 .5 L$
$∵$ $1 c m^{3} = 1 m L = 1000 L$
$∴ M o l a r i t y$
$= \frac{M a s s o f s o l u t e i n g r a m s (w)}{M o l a r m a s s o f s o l u t e (M) \times V o l u m e o f s o l u t i o n V (L))}$
$M o l a r i t y o f u r e a = \frac{15 g}{60 g m o l^{- 1} \times 0 .5 L}$
$\Rightarrow 0 .5 m o l / L = 0 .5 m o l / d m^{3}$
$(1 L = 1 d m^{3})$

MHTCET - 2018

CHXII02:SOLUTIONS

319210 Three statements are given about mole fraction
i. Mole fraction of a solute + mole fraction of solvent = 1
ii. Equal weights of Helium and methane are present in a gaseous mixture. The mole fraction of He is 4/5
iii. The mole fraction of $40 %$ by mass of NaOH solution is 0.5

1 i and ii are correct

2 ii and iii are correct

3 i and iii are correct

4 all are correct

Explanation:

(i) Sum of the mole fraction of all the components in solution is 1.
(ii) Let us assume weight of $H E a n d C H_{4}$ as 1 g Mole fraction of $H E = \frac{\frac{1}{4}}{\frac{1}{4} + \frac{1}{6}} = \frac{4}{5}$ .
(iii) 40% by mass of NaOH means 40 g of NaOH present in 100 g of solution.
Mole fraction of $N a O H = \frac{\frac{40}{40}}{(\frac{40}{40}) + (\frac{60}{18})} = 0 .23$

CHXII02:SOLUTIONS

319207 The molarity of a solution obtained by mixing of 750 mL of 0.5 M HCl with 250 mL of 2 M HCl will be

1 0.875 M

2 1 M

3 1.25 M

4 2.5 M

Explanation:

$M_{M i x t u r e} = \frac{M_{1} V_{1} + M_{2} V_{2}}{V_{1} + V_{2}}$
$= \frac{0 .5 \times 750 + 2 \times 250}{750 + 250} = 0 .875 M$

CHXII02:SOLUTIONS

319208 A student has 100 mL of 0.1 M KCl solution. To make it 0.2 M, he has to

1 evaporate 50 mL of the solution

2 add 0.01 mole of KCl

3 both (1) and (2) can be used

4 neither (1) nor (2) can be used

Explanation:

100 mL of 0.1 M KCl contains 0.01 mole of KCl
On evaporating 50 mL of solution, now conc.
$= \frac{0 .01}{50} \times 1000 = 0 .2 M$
On adding 0.01 mole of KCl, now KCl present = 0.02 mole
$∴$ Concentration $= \frac{0 .02}{100} \times 1000 = 0 .2 M$
Thus, both (1) and (2) can be used.

CHXII02:SOLUTIONS

319209 The molarity of urea (molar mass $60 g m o l^{- 1}$ ) solution by dissolving 15 g of urea in $500 c m^{3}$ of water is

1

2 m o l d m^{- 3}

2

0 .5 m o l d m^{- 3}

3

0 .125 m o l d m^{- 3}

4

0 .0005 m o l d m^{- 3}

Explanation:

Given, mass of urea = 15 g
Molar mass of urea $= 60 g m o l^{- 1}$
$V o l u m e = 500 c m^{3} = 500 m L = 0 .5 L$
$∵$ $1 c m^{3} = 1 m L = 1000 L$
$∴ M o l a r i t y$
$= \frac{M a s s o f s o l u t e i n g r a m s (w)}{M o l a r m a s s o f s o l u t e (M) \times V o l u m e o f s o l u t i o n V (L))}$
$M o l a r i t y o f u r e a = \frac{15 g}{60 g m o l^{- 1} \times 0 .5 L}$
$\Rightarrow 0 .5 m o l / L = 0 .5 m o l / d m^{3}$
$(1 L = 1 d m^{3})$

MHTCET - 2018

CHXII02:SOLUTIONS

319210 Three statements are given about mole fraction
i. Mole fraction of a solute + mole fraction of solvent = 1
ii. Equal weights of Helium and methane are present in a gaseous mixture. The mole fraction of He is 4/5
iii. The mole fraction of $40 %$ by mass of NaOH solution is 0.5

1 i and ii are correct

2 ii and iii are correct

3 i and iii are correct

4 all are correct

Explanation:

(i) Sum of the mole fraction of all the components in solution is 1.
(ii) Let us assume weight of $H E a n d C H_{4}$ as 1 g Mole fraction of $H E = \frac{\frac{1}{4}}{\frac{1}{4} + \frac{1}{6}} = \frac{4}{5}$ .
(iii) 40% by mass of NaOH means 40 g of NaOH present in 100 g of solution.
Mole fraction of $N a O H = \frac{\frac{40}{40}}{(\frac{40}{40}) + (\frac{60}{18})} = 0 .23$