150 ~g of acetic acid was contaminated with 10.2 ~g ascorbic acid (C_6 H_8 O_6) to lower down its freezing point by (x \times 10^-1)^\circ C . The value of x is \underline\hspace2cm . (Nearest integer) [Given K_f=3.9 ~K ~kg ~mol^-1 ; molar mass of ascorbic acid =176 ~g ~mol^-1 ]

Solutions PYQ — JEE Main 2022 · 28 Jul · Shift 1 · Q16 with Solution

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Correct answer: 15

For a non-electrolyte solute,

\Delta T_f = K_f m

where $m$ is molality.

Given mass of ascorbic acid $=10.2\,\text{g}$

Molar mass $=176\,\text{g mol}^{-1}$

So,

\text{moles of solute} = \frac{10.2}{176}

=0.05795\,\text{mol}

Mass of acetic acid $=150\,\text{g}=0.150\,\text{kg}$

m = \frac{0.05795}{0.150} = 0.3863\,\text{mol kg}^{-1}

\Delta T_f = 3.9 \times 0.3863

\Delta T_f \approx 1.506\,\text{K}

Since temperature difference in K and $^\circ$ C are numerically same,

\Delta T_f \approx 1.5^\circ \text{C}

Given,

\Delta T_f = (x \times 10^{-1})^\circ \text{C}

So,

x \times 10^{-1} = 1.5

x = 15

\boxed{15}