New chat PYQ Library Doubts Rank report

Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →

Solutions question

2017 · Shift 0 · Q19

Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.

PYQ Library
JEE Main
Chemistry
Solutions
2017 · Shift 0 · Q19

Solutions question

2017 · Shift 0 · Q19

JEE MainChemistrySolutionsMCQ+4 / −1

The freezing point of benzene decreases by 0.450C when 0.2 g of acetic acid is added to 20g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be: (Kf for benzene = 5.12 K kg mol–1)

A
80.4 %
B
74.6 %
C
94.6 %
D
64.6 %

View written solutionFree

Correct answer: C

Use freezing point depression formula

For a solute in a solvent,

\Delta T_f = i K_f m

where:

$\Delta T_f = 0.45\,\text{K}$
$K_f = 5.12\,\text{K kg mol}^{-1}$
$i$ is the van't Hoff factor
$m$ is molality

Calculate molality assuming no association

Mass of acetic acid $=0.2\,\text{g}$

Molar mass of acetic acid $(\text{CH}_3\text{COOH}) = 60\,\text{g mol}^{-1}$

So moles of acetic acid:

n = \frac{0.2}{60} = 0.003333\,\text{mol}

Mass of benzene $=20\,\text{g}=0.020\,\text{kg}$

Thus molality:

m = \frac{0.003333}{0.020} = 0.1667\,\text{mol kg}^{-1}

Find van't Hoff factor

From

0.45 = i \times 5.12 \times 0.1667

First compute:

5.12 \times 0.1667 \approx 0.8533

Hence,

i = \frac{0.45}{0.8533} \approx 0.527

Relate $i$ to degree of association

Acetic acid associates as:

2A \rightarrow A_2

If degree of association is $\alpha$ , then for dimerization,

i = 1 - \frac{\alpha}{2}

So,

0.527 = 1 - \frac{\alpha}{2}

\frac{\alpha}{2} = 1 - 0.527 = 0.473

\alpha = 0.946

Therefore percentage association is:

0.946 \times 100 = 94.6\%

Check options

A: $80.4\%$ ❌
B: $74.6\%$ ❌
C: $94.6\%$ ✅
D: $64.6\%$ ❌

So the correct option is C.

More from Solutions

The solubility of $N_{2}$ in water at 300 K and 500 torr partial pressure is 0.01 g L−1. The solubility (in g L−1) at 750 torr partial pressure is :2016 · MCQ
An aqueous solution of a salt $M X_{2}$ at certain temperature has a van’t Hoff factor of 2. The degree of dissociation for this solution of the salt is :2016 · MCQ
18 g glucose ( $C_{6} H_{12} O_{6}$ ) is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous solution is :2016 · MCQ
The vapour pressure of acetone at 20oC is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20oC, its vapour pressure was 183 torr. The molar mass (g mol-1) of the substance is:2015 · MCQ
Consider separate solutions of 0.500 M $C_{2} H_{5} O H$ (aq), 0.100 M $M g_{3} (P O_{4})_{2}$ (aq), 0.250 M $K B r$ (aq) and 0.125 M $N a_{3} P O_{4}$ (aq) at 25oC. Which statement is true about these solutions, assuming all salts to be strong electrolytes?2014 · MCQ
Kf for water is 1.86K kg mol–1. If your automobile radiator holds 1.0 kg of water, how many grams of ethylene glycol ( $C_{2} H_{6} O_{2}$ ) must you add to get the freezing point of the solution lowered to –2.8oC ?2012 · MCQ
The degree of dissociation ( $α$ ) of a weak electrolyte, AxBy is related to van’t Hoff factor (i) by the expression :2011 · MCQ
Ethylene glycol is used as an antifreeze in a cold climate. Mass of ethylene glycol which should be added to 4 kg of water to prevent it from freezing at −6oC will be : [Kf for water = 1.86 K kg mol−1 , and molar mass of ethylene glycol =…2011 · MCQ