QBManager

CHXII02:SOLUTIONS

319211 Which of the following is correct?

1 For a binary solution sum of the mole fractions of all components is equal to one

2

\frac{M o l e f r a c t i o n o f I}{M o l e f r a c t i o n o f I I} = \frac{m o l e o f I}{m o l e o f I I}

(f o r b i n a r y s o l u t i o n)

3 Mole fraction of solute

= \frac{m o l e s o f s o l u t e}{m o l e s o f s o l u t e + m o l e s o f s o l v e n t}

4 All are correct

CHXII02:SOLUTIONS

319212 What is the mole fraction of glucose in $10 % w /$ w glucose solution?

1 0.01

2 0.02

3 0.03

4 0.04

Explanation:

No. of moles of glucose $= \frac{10}{180} = 0.0555$
No. of moles of water $= \frac{90}{18} = 5$
$∴$ No. of moles of solution $= 5.0555$
Mole fraction of glucose $=$
$\frac{No . of moles of glucose}{No . of moles of solution}$
$= \frac{0.0555}{5.0555} = 0.01$

CHXII02:SOLUTIONS

319213 $138 g$ ethyl alcohol is mixed with $72 g$ of water. The ratio of mole fraction of alcohol to water is

1

3 : 4

2

1 : 2

3

1 : 4

4

1 : 1

Explanation:

Number of moles of $C_{2} H_{5} OH$
$= \frac{138}{46} = 3 (M_{C_{2} H_{5} O H} = 46 g / m o l)$
Number of moles of $H_{2} O = \frac{72}{18} = 4$
$(∵ M_{H_{2} 0} = 18 g / mol)$
Mole fraction, $χ_{C_{2} H_{5} OH} = \frac{3}{3 + 4} = \frac{3}{7}$
$χ_{H_{2} O} = \frac{4}{3 + 4} = \frac{4}{7}$
$\frac{χ_{C_{2} H_{5} O H}}{χ_{H_{2} O}} = \frac{3 / 7}{4 / 7}$
$= 3 / 4 = 3 : 4$

CHXII02:SOLUTIONS

319214 Assertion :
$1.575 g H_{2} C_{2} O_{4} \cdot 2 H_{2} O$ in $250 mL$ solution makes it $0.1 N$ .
Reason :
$H_{2} C_{2} O_{4} \cdot 2 H_{2} O$ is a dihydrate organic acid.

1 If both Assertion & Reason are true and the reason is the correct explanation of the assertion.

2 If both Assertion & Reason are true but the reason is not the correct explanation of the assertion.

3 If Assertion is true but Reason is false.

4 Both Assertion and Reason are false

Explanation:

Normality $= \frac{mass}{eq.mass} \times \frac{1000}{V (mL)}$
(eq.mass of $H_{2} C_{2} O_{4} \cdot 2 H_{2} O$ is 63 as it is a dibasic acid)
$N = \frac{1.575}{63} \times \frac{1000}{250} = 0.1 N$
$H_{2} C_{2} O_{4} \cdot 2 H_{2} O$ is a dihydrate organic acid, since it contains 2 molecules of water of crystallisation. But it is not the required explanation.

AIIMS - 2009

CHXII02:SOLUTIONS

319211 Which of the following is correct?

1 For a binary solution sum of the mole fractions of all components is equal to one

2

\frac{M o l e f r a c t i o n o f I}{M o l e f r a c t i o n o f I I} = \frac{m o l e o f I}{m o l e o f I I}

(f o r b i n a r y s o l u t i o n)

3 Mole fraction of solute

= \frac{m o l e s o f s o l u t e}{m o l e s o f s o l u t e + m o l e s o f s o l v e n t}

4 All are correct

CHXII02:SOLUTIONS

319212 What is the mole fraction of glucose in $10 % w /$ w glucose solution?

1 0.01

2 0.02

3 0.03

4 0.04

Explanation:

No. of moles of glucose $= \frac{10}{180} = 0.0555$
No. of moles of water $= \frac{90}{18} = 5$
$∴$ No. of moles of solution $= 5.0555$
Mole fraction of glucose $=$
$\frac{No . of moles of glucose}{No . of moles of solution}$
$= \frac{0.0555}{5.0555} = 0.01$

CHXII02:SOLUTIONS

319213 $138 g$ ethyl alcohol is mixed with $72 g$ of water. The ratio of mole fraction of alcohol to water is

1

3 : 4

2

1 : 2

3

1 : 4

4

1 : 1

Explanation:

Number of moles of $C_{2} H_{5} OH$
$= \frac{138}{46} = 3 (M_{C_{2} H_{5} O H} = 46 g / m o l)$
Number of moles of $H_{2} O = \frac{72}{18} = 4$
$(∵ M_{H_{2} 0} = 18 g / mol)$
Mole fraction, $χ_{C_{2} H_{5} OH} = \frac{3}{3 + 4} = \frac{3}{7}$
$χ_{H_{2} O} = \frac{4}{3 + 4} = \frac{4}{7}$
$\frac{χ_{C_{2} H_{5} O H}}{χ_{H_{2} O}} = \frac{3 / 7}{4 / 7}$
$= 3 / 4 = 3 : 4$

CHXII02:SOLUTIONS

319214 Assertion :
$1.575 g H_{2} C_{2} O_{4} \cdot 2 H_{2} O$ in $250 mL$ solution makes it $0.1 N$ .
Reason :
$H_{2} C_{2} O_{4} \cdot 2 H_{2} O$ is a dihydrate organic acid.

1 If both Assertion & Reason are true and the reason is the correct explanation of the assertion.

2 If both Assertion & Reason are true but the reason is not the correct explanation of the assertion.

3 If Assertion is true but Reason is false.

4 Both Assertion and Reason are false

Explanation:

Normality $= \frac{mass}{eq.mass} \times \frac{1000}{V (mL)}$
(eq.mass of $H_{2} C_{2} O_{4} \cdot 2 H_{2} O$ is 63 as it is a dibasic acid)
$N = \frac{1.575}{63} \times \frac{1000}{250} = 0.1 N$
$H_{2} C_{2} O_{4} \cdot 2 H_{2} O$ is a dihydrate organic acid, since it contains 2 molecules of water of crystallisation. But it is not the required explanation.

AIIMS - 2009

CHXII02:SOLUTIONS

319211 Which of the following is correct?

1 For a binary solution sum of the mole fractions of all components is equal to one

2

\frac{M o l e f r a c t i o n o f I}{M o l e f r a c t i o n o f I I} = \frac{m o l e o f I}{m o l e o f I I}

(f o r b i n a r y s o l u t i o n)

3 Mole fraction of solute

= \frac{m o l e s o f s o l u t e}{m o l e s o f s o l u t e + m o l e s o f s o l v e n t}

4 All are correct

CHXII02:SOLUTIONS

319212 What is the mole fraction of glucose in $10 % w /$ w glucose solution?

1 0.01

2 0.02

3 0.03

4 0.04

Explanation:

No. of moles of glucose $= \frac{10}{180} = 0.0555$
No. of moles of water $= \frac{90}{18} = 5$
$∴$ No. of moles of solution $= 5.0555$
Mole fraction of glucose $=$
$\frac{No . of moles of glucose}{No . of moles of solution}$
$= \frac{0.0555}{5.0555} = 0.01$

CHXII02:SOLUTIONS

319213 $138 g$ ethyl alcohol is mixed with $72 g$ of water. The ratio of mole fraction of alcohol to water is

1

3 : 4

2

1 : 2

3

1 : 4

4

1 : 1

Explanation:

Number of moles of $C_{2} H_{5} OH$
$= \frac{138}{46} = 3 (M_{C_{2} H_{5} O H} = 46 g / m o l)$
Number of moles of $H_{2} O = \frac{72}{18} = 4$
$(∵ M_{H_{2} 0} = 18 g / mol)$
Mole fraction, $χ_{C_{2} H_{5} OH} = \frac{3}{3 + 4} = \frac{3}{7}$
$χ_{H_{2} O} = \frac{4}{3 + 4} = \frac{4}{7}$
$\frac{χ_{C_{2} H_{5} O H}}{χ_{H_{2} O}} = \frac{3 / 7}{4 / 7}$
$= 3 / 4 = 3 : 4$

CHXII02:SOLUTIONS

319214 Assertion :
$1.575 g H_{2} C_{2} O_{4} \cdot 2 H_{2} O$ in $250 mL$ solution makes it $0.1 N$ .
Reason :
$H_{2} C_{2} O_{4} \cdot 2 H_{2} O$ is a dihydrate organic acid.

1 If both Assertion & Reason are true and the reason is the correct explanation of the assertion.

2 If both Assertion & Reason are true but the reason is not the correct explanation of the assertion.

3 If Assertion is true but Reason is false.

4 Both Assertion and Reason are false

Explanation:

Normality $= \frac{mass}{eq.mass} \times \frac{1000}{V (mL)}$
(eq.mass of $H_{2} C_{2} O_{4} \cdot 2 H_{2} O$ is 63 as it is a dibasic acid)
$N = \frac{1.575}{63} \times \frac{1000}{250} = 0.1 N$
$H_{2} C_{2} O_{4} \cdot 2 H_{2} O$ is a dihydrate organic acid, since it contains 2 molecules of water of crystallisation. But it is not the required explanation.

AIIMS - 2009

CHXII02:SOLUTIONS

319211 Which of the following is correct?

1 For a binary solution sum of the mole fractions of all components is equal to one

2

\frac{M o l e f r a c t i o n o f I}{M o l e f r a c t i o n o f I I} = \frac{m o l e o f I}{m o l e o f I I}

(f o r b i n a r y s o l u t i o n)

3 Mole fraction of solute

= \frac{m o l e s o f s o l u t e}{m o l e s o f s o l u t e + m o l e s o f s o l v e n t}

4 All are correct

CHXII02:SOLUTIONS

319212 What is the mole fraction of glucose in $10 % w /$ w glucose solution?

1 0.01

2 0.02

3 0.03

4 0.04

Explanation:

No. of moles of glucose $= \frac{10}{180} = 0.0555$
No. of moles of water $= \frac{90}{18} = 5$
$∴$ No. of moles of solution $= 5.0555$
Mole fraction of glucose $=$
$\frac{No . of moles of glucose}{No . of moles of solution}$
$= \frac{0.0555}{5.0555} = 0.01$

CHXII02:SOLUTIONS

319213 $138 g$ ethyl alcohol is mixed with $72 g$ of water. The ratio of mole fraction of alcohol to water is

1

3 : 4

2

1 : 2

3

1 : 4

4

1 : 1

Explanation:

Number of moles of $C_{2} H_{5} OH$
$= \frac{138}{46} = 3 (M_{C_{2} H_{5} O H} = 46 g / m o l)$
Number of moles of $H_{2} O = \frac{72}{18} = 4$
$(∵ M_{H_{2} 0} = 18 g / mol)$
Mole fraction, $χ_{C_{2} H_{5} OH} = \frac{3}{3 + 4} = \frac{3}{7}$
$χ_{H_{2} O} = \frac{4}{3 + 4} = \frac{4}{7}$
$\frac{χ_{C_{2} H_{5} O H}}{χ_{H_{2} O}} = \frac{3 / 7}{4 / 7}$
$= 3 / 4 = 3 : 4$

CHXII02:SOLUTIONS

319214 Assertion :
$1.575 g H_{2} C_{2} O_{4} \cdot 2 H_{2} O$ in $250 mL$ solution makes it $0.1 N$ .
Reason :
$H_{2} C_{2} O_{4} \cdot 2 H_{2} O$ is a dihydrate organic acid.

1 If both Assertion & Reason are true and the reason is the correct explanation of the assertion.

2 If both Assertion & Reason are true but the reason is not the correct explanation of the assertion.

3 If Assertion is true but Reason is false.

4 Both Assertion and Reason are false

Explanation:

Normality $= \frac{mass}{eq.mass} \times \frac{1000}{V (mL)}$
(eq.mass of $H_{2} C_{2} O_{4} \cdot 2 H_{2} O$ is 63 as it is a dibasic acid)
$N = \frac{1.575}{63} \times \frac{1000}{250} = 0.1 N$
$H_{2} C_{2} O_{4} \cdot 2 H_{2} O$ is a dihydrate organic acid, since it contains 2 molecules of water of crystallisation. But it is not the required explanation.

AIIMS - 2009

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation: