When ' x ' × 10 -2 mL methanol (molar mass =32 g ; density =0.792 g / cm 3 ) is added to 100 mL water ( . density .=1 g / cm 3 ) , the following diagram is obtained. [Given: Molal freezing point depression constant of water at 273.15 K -1 is 1.86 K kg mol -1 ]

Chemistry · Solutions and Colligative Properties

JEE Main 2024 — 6 April, Shift 2 — Question 79

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 mL water $\left(\right.$ density $\left.=1 \mathrm{~g} / \mathrm{cm}^{3}\right)$ , the following diagram is obtained.

[Given: Molal freezing point depression constant of water at $273.15 \mathrm{~K}^{-1}$ is $1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ ]

Answer: 543

Numerical answer — enter this value.

Step-by-step solution

$\Delta \mathrm{T}_{\mathrm{f}}=273.15-270.65=2.5 \mathrm{~K}$

$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \mathrm{m} \Rightarrow 2.5=1.86 \times \frac{\mathrm{n}}{0.1}$ $\Rightarrow \mathrm{n}=0.1344$ moles

$\Rightarrow \mathrm{w}=0.1344 \times 32=4.3 \mathrm{~g}$

Volume $=\frac{4.3}{0.792}=5.43 \mathrm{ml}=543 \times 10^{-2} \mathrm{ml}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 6 April, Shift 2
Subject: Chemistry
Chapter: Solutions and Colligative Properties
Topic: Solid in Liquid Solutions (Colligative Properties)