JEE MAIN - Chemistry (2024 - 6th April Evening Shift - No. 23)
When '$$x$$' $$\times 10^{-2} \mathrm{~mL}$$ methanol (molar mass $$=32 \mathrm{~g}$$' density $$=0.792 \mathrm{~g} / \mathrm{cm}^3$$) is added to $$100 \mathrm{~mL}$$. water (density $$=1 \mathrm{~g} / \mathrm{cm}^3$$), the following diagram is obtained.
_6th_April_Evening_Shift_en_23_1.png)
$$x=$$ ________ (nearest integer).
[Given : Molal freezing point depression constant of water at $$273.15 \mathrm{~K}$$ is $$1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$$]
Answer
543
Explanation
$$\begin{aligned}
& \Delta T_f=2.5^{\circ} \mathrm{C} \\
& \Delta T_f=i \times k_f \times m \\
& 2.5=1 \times 1.86 \times \frac{n_B}{0.1} \\
& n_B=\frac{2.5 \times 0.1}{1.86}=0.1344 \mathrm{~mol} \\
& \text { mass of methanol }=0.1344 \times 32=4.3 \mathrm{~g} \\
& d=\frac{m}{v} \\
& v=\frac{m}{d} \\
& v=\frac{4.3}{0.792} \mathrm{~mL} \\
& =5.43 \mathrm{~mL} \\
& =543 \times 10^{-2} \mathrm{~mL} \\
& x=543
\end{aligned}$$
Comments (0)
