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JEE Main 20246 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2024 · 6 April · Evening Shift · Chemistry · Solutions

When ' x ' ( × 10^ -2 ~mL ) methanol (molar mass (=32 ~g ); density...

When ' xx ' (×\times10^{-2}  mL\mathrm{~mL}) methanol (molar mass (=32  g\mathrm{~g}); density (=0.792  g\mathrm{~g}/ \mathrm{cm}^3)) is added to (100  mL\mathrm{~mL}) water (density (=1  g\mathrm{~g}/ \mathrm{cm}^3)), the following diagram is obtained. When ' x ' ( × 10^ -2 ~mL ) methanol (molar mass (=32 ~g ); density (=0.792 ~g / cm^ 3)) is added to (100 ~mL ) water
x=x= ______ (nearest integer). [Given : Molal freezing point depression constant of water at (273.15  K\mathrm{~K}) is (1.86  K kg mol−1\mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1})]

Correct answer

0

Step-by-step solution

ΔTf=273.15−270.65=2.5KΔTf=Kfm⇒2.5=1.86×n0.1⇒n=0.1344 moles ⇒w=0.1344×32=4.3g Volume =4.30.792=5.43ml=543×10−2ml\begin{aligned} & \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 \text{ moles } \\ & \Rightarrow \mathrm{w}=0.1344 \times 32=4.3 \mathrm{ g} \\ & \text{ Volume }=\frac{4.3}{0.792}=5.43 \mathrm{ml}=543 \times 10^{-2} \mathrm{ml}\end{aligned}
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