JEE MAIN - Chemistry (2022 - 28th July Morning Shift - No. 16)
$$150 \mathrm{~g}$$ 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
15
Explanation
M.wt. of Acetic acid $=60 \mathrm{~g}$
M.wt. of Ascorbic acid $=176 \mathrm{~g}$
$$ \Delta T_{f}=K_{f} m $$
$$ \begin{aligned} \Delta T_{f} &=\frac{3.9 \times 10.2 \times 1000}{176 \times 150} \\\\ \Delta T_{f} &=1.506 \\\\ &=15.06 \times 10^{-1} \\\\ &=15 \end{aligned} $$
M.wt. of Ascorbic acid $=176 \mathrm{~g}$
$$ \Delta T_{f}=K_{f} m $$
$$ \begin{aligned} \Delta T_{f} &=\frac{3.9 \times 10.2 \times 1000}{176 \times 150} \\\\ \Delta T_{f} &=1.506 \\\\ &=15.06 \times 10^{-1} \\\\ &=15 \end{aligned} $$
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