$150 \mathrm{~g}$ of acetic acid was contaminated with $10.2 \mathrm{~g}$ ascorbic acid $\left(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}\right)$ to lower down its freezing point by $\left(x \times 10^{-1}\right)^{\circ} \mathrm{C}$. The value of $x$ is ___________. (Nearest integer)
[Given $\mathrm{K}_{f}=3.9 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$; molar mass of ascorbic acid $=176 \mathrm{~g} \mathrm{~mol}^{-1}$]
Answer (integer)
15
Solution
M.wt. of Acetic acid $=60 \mathrm{~g}$
<br/><br/>
M.wt. of Ascorbic acid $=176 \mathrm{~g}$
<br/><br/>
$\Delta T_{f}=K_{f} m$
<br/><br/>
$$
\begin{aligned}
\Delta T_{f} &=\frac{3.9 \times 10.2 \times 1000}{176 \times 150} \\\\
\Delta T_{f} &=1.506 \\\\
&=15.06 \times 10^{-1} \\\\
&=15
\end{aligned}
$$
About this question
Subject: Chemistry · Chapter: States of Matter · Topic: Gas Laws
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