The resistivity of a $0.8 \mathrm{M}$ solution of an electrolyte is $5 \times 10^{-3} \Omega~ \mathrm{cm}$.
Its molar conductivity is _________ $\times 10^{4}~ \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$. (Nearest integer)
Answer (integer)
25
Solution
$$
\begin{aligned}
\text { Molar conductivity } & =\frac{\mathrm{k} \times 1000}{\mathrm{C}} \\\\
& =\frac{\frac{1}{5 \times 10^{-3}} \times 1000}{0.8} \\\\
& =\frac{10^6}{4}=0.25 \times 10^6 \\\\
& =25 \times 10^4 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}
\end{aligned}
$$
About this question
Subject: Chemistry · Chapter: Electrochemistry · Topic: Electrochemical Cells
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