0.2 \%( w / v ) solution of NaOH is measured to have resistivity 870.0 m Ω m . The molar conductivity of the solution will be \ \ \ \ \ × 10 2 mS dm 2 mol -1 . (Nearest integer)

Chemistry · Electrochemistry

JEE Main 2025 — 2 April, Evening Shift — Question 19

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

Answer: 23

Numerical answer — enter this value.

Step-by-step solution

We have 0.2 g of NaOH in 100 mL solution

\begin{aligned} & M=\frac{0.2}{40 \times 100} \times 1000 \\& =0.05 \mathrm{M} \\& \kappa=\frac{1}{0.870}=1.15 \mathrm{sm}^{-1} \\& =1.15 \times 10^{-2} \mathrm{~S} \mathrm{~cm}^{-1} \\& \wedge_{\mathrm{m}}=\frac{\kappa \times 1000}{\mathrm{~m}}=\frac{1.15 \times 10^{-2} \times 1000}{0.05} \\& =230 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \\& =23 \times 10^{4} \mathrm{~m} \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \\& =23 \times 10^{2} \mathrm{~m} \mathrm{~S} \mathrm{dm}^{2} \mathrm{~mol}^{-1} \end{aligned}

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Exam: JEE Main 2025
Paper: 2 April, Evening Shift
Subject: Chemistry
Chapter: Electrochemistry
Topic: Conductance of Solutions and Kohlrausch's Law