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Solutions question

2019 · 12 Jan · Shift 2 · Q19

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2019 · 12 Jan · Shift 2 · Q19

Solutions question

2019 · 12 Jan · Shift 2 · Q19

JEE MainChemistrySolutionsMCQ+4 / −1

Molecules of benzoic acid (

C_6H_5COOH

) dimerise in benzene. 'w' g of the acid dissolved in 30 g of benzene shows a depression in freezing point equal to 2K. If the percentage association of the acid to form dimmer in the solution is 80, then w is – (Its given that Kf = 5 K kg mol–1, Molar mass of benzoic acid = 122 g mol–1)

A
1.5 g
B
1.8 g
C
1.0 g
D
2.4 g

View written solutionFree

Correct answer: D

Use freezing point depression formula

For a solute showing association/dissociation,

\Delta T_f = i K_f m

where:

$\Delta T_f = 2\,\text{K}$
$K_f = 5\,\text{K kg mol}^{-1}$
$i$ = van’t Hoff factor
$m$ = molality

So,

2 = i \cdot 5 \cdot m

Find van’t Hoff factor for dimerisation

Benzoic acid dimerises:

2A \rightarrow A_2

If degree of association is $\alpha = 80\% = 0.8$ , then starting from 1 mole of $A$ :

unassociated moles = $1-\alpha = 0.2$
dimer formed = $\alpha/2 = 0.4$

Total moles in solution:

0.2 + 0.4 = 0.6

Hence,

i = 0.6

Equivalently, for dimerisation:

i = 1-\frac{\alpha}{2} = 1-\frac{0.8}{2} = 0.6

Calculate molality

From

2 = 0.6 \times 5 \times m

m = \frac{2}{3} \text{ mol kg}^{-1}

Use definition of molality

Mass of benzene = $30\,\text{g} = 0.03\,\text{kg}$

If $n$ is moles of benzoic acid taken, then

m = \frac{n}{0.03}

So,

\frac{n}{0.03} = \frac{2}{3}

n = \frac{2}{3}\times 0.03 = 0.02\,\text{mol}

Find mass of benzoic acid

Molar mass of benzoic acid = $122\,\text{g mol}^{-1}$

w = nM = 0.02 \times 122 = 2.44\,\text{g}

Thus,

w \approx 2.4\,\text{g}

Check options

A: $1.5\,\text{g}$ — incorrect
B: $1.8\,\text{g}$ — incorrect
C: $1.0\,\text{g}$ — incorrect
D: $2.4\,\text{g}$ — correct

Therefore, the correct option is D.

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