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JEE Main 202227 JuneEvening Shift
Exam
JEE Main
Year
2022
Date
27 June
Shift
Evening Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2022 · 27 June · Evening Shift · Chemistry · Solutions

A solution containing 2 . 5 × 10 ^ - 3 kg of a solute dissolved in...

A solution containing 2.5×10−3kg2 . 5 \times {10}^{- 3}kg of a solute dissolved in 75×10−3kg75 \times {10}^{- 3}kg of water boils at 373.535K373 . 535K. The molar mass of the solute is mol−1⋅{mol}^{- 1} \cdot [nearest integer] (Given : Kb(H2O)=0.52Kkgmol−1{K}_{b}\left({H}_{2}O\right) = 0 . 52Kkg{mol}^{- 1} and boiling point of water =373.15K= 373 . 15K)

Correct answer

0

Step-by-step solution

ΔTb=iKbm{\Delta T}_{b} = {iK}_{b}m

ΔTb=ElevationofboilingpointKb=molalboilingpointelevationconstantm=molalityi=van′thofffactor{\Delta T}_{b} = Elevationofboilingpoint{K}_{b} = molalboilingpointelevationconstantm = molalityi = van ' thofffactor

m=2.5M×100075m = \dfrac{2 . 5}{M} \times \dfrac{1000}{75}

∆Tb=Tb−TboTb=BoilingpointofthesolutionTbo=Boilingpointofthesolvent∆Tb=373.535−373.15=0.385K∆ {T}_{b} = {T}_{b} - {T}_{b}^{o}{T}_{b} = Boilingpointofthesolution{T}_{b}^{o} = Boilingpointofthesolvent ∆ {T}_{b} = 373 . 535 - 373 . 15 = 0 . 385K

ΔTb=iKbm{\Delta T}_{b} = {iK}_{b}m

0.385=1×0.52×2.5×1000M×750 . 385 = 1 \times 0 . 52 \times \dfrac{2 . 5 \times 1000}{M \times 75}

The molar mass of the solute,M=45g/mlM = 45g / ml.

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