JEE MAIN - Chemistry (2024 - 4th April Evening Shift - No. 25)
$$2.7 \mathrm{~kg}$$ of each of water and acetic acid are mixed. The freezing point of the solution will be $$-x^{\circ} \mathrm{C}$$. Consider the acetic acid does not dimerise in water, nor dissociates in water. $$x=$$ ________ (nearest integer)
[Given: Molar mass of water $$=18 \mathrm{~g} \mathrm{~mol}^{-1}$$, acetic acid $$=60 \mathrm{~g} \mathrm{~mol}^{-1}$$
$${ }^{\mathrm{K}_{\mathrm{f}}} \mathrm{H}_2 \mathrm{O}: 1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$$
$$\mathrm{K}_{\mathrm{f}}$$ acetic acid: $$3.90 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$$
freezing point: $$\mathrm{H}_2 \mathrm{O}=273 \mathrm{~K}$$, acetic acid $$=290 \mathrm{~K}$$]
Answer
31
Explanation
$$\begin{aligned}
& \text { Molality of acetic acid }=\frac{2700}{60} \times \frac{1}{2.7} \mathrm{~mol} / \mathrm{kg} \\
&=16.667 \\
& \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \times 16.667 \\
&=1.86 \times 16.667 \\
&=31 \mathrm{~K}
\end{aligned}$$
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