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JEE Main 20252 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Electrochemistry

JEE Main 2025 · 2 April · Evening Shift · Chemistry · Electrochemistry

0.2 \%(w / v) solution of NaOH is measured to have resistivity...

0.2%(w/v)0.2 \%(\mathrm{w} / \mathrm{v}) solution of NaOH is measured to have resistivity 870.0 mΩ m870.0 \mathrm{~m} \Omega \mathrm{~m}. The molar conductivity of the solution will be ______ ×102mSdm2 mol−1\times 10^2 \mathrm{mS} \mathrm{dm}{ }^2 \mathrm{~mol}^{-1}. (Nearest integer)

Correct answer

0

Step-by-step solution

Given : Concentration of NaOH=0.2%(w/v)\mathrm{NaOH}=0.2 \%(\mathrm{w} / \mathrm{v})
∴0.2 g\therefore 0.2 \mathrm{~g} of NaOH in 100 ml of solution.
Molarity of NaOH solution
<br/>= moles of solute Vml×1000<br>=0.2/40100×1000==0.240×100×1000=240M<br/>\begin{array}{r}<br />=\frac{\text{ moles of solute }}{\mathrm{V}_{\mathrm{ml}}} \times 1000 \\<br>=\frac{0.2 / 40}{100} \times 1000==\frac{0.2}{40 \times 100} \times 1000=\frac{2}{40} \mathrm{M}<br />\end{array}
Given resistivity of solution =870 m=870 \mathrm{~m} ohm m =870×10−3ohmm=870 \times 10^{-3} \mathrm{ohm} \mathrm{m}
=870×10−3×10ohmdm=870 \times 10^{-3} \times 10 \mathrm{ohm} \mathrm{dm}
=870×10−2=870 \times 10^{-2} ohm dm
=8.7=8.7 ohm dm
Now conductivity
K=1ρ=18.7ohm−1dm−1\mathrm{K}=\frac{1}{\rho}=\frac{1}{8.7} \mathrm{ohm}^{-1} \mathrm{dm}^{-1}
Now molar conductivity of solution is
<br/>λm=KM=18.7240=402×8.7=2.29 Sdm2 mol−1<br>2.29×103 m Sdm2 mol−1<br>=22.9×102 m Sdm2 mol−1<br>=23×102 m Sdm2 mol−1<br/>\begin{aligned}<br />& \lambda_{\mathrm{m}}=\frac{\mathrm{K}}{\mathrm{M}}=\frac{\frac{1}{8.7}}{\frac{2}{40}}=\frac{40}{2 \times 8.7}=2.29 \mathrm{~S} \mathrm{dm}^2 \mathrm{~mol}^{-1} \\<br>& 2.29 \times 10^3 \mathrm{~m} \mathrm{~S} \mathrm{dm}^2 \mathrm{~mol}^{-1} \\<br>& =22.9 \times 10^2 \mathrm{~m} \mathrm{~S} \mathrm{dm}^2 \mathrm{~mol}^{-1} \\<br>& =23 \times 10^2 \mathrm{~m} \mathrm{~S} \mathrm{dm}^2 \mathrm{~mol}^{-1}<br />\end{aligned}
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