Resistance of a conductivity cell filled with a solution of an electrolyte of concentration $0.1 \mathrm{M}$…

Resistance of a conductivity cell filled with a solution of an electrolyte of concentration $0.1 \mathrm{M}$ is $100 \Omega$. The conductivity of this solution is $1.29 \mathrm{~S} \mathrm{~m}^{-1}$. Resistance of the same cell when filled with $0.2 \mathrm{M}$ of the same solution is $520 \Omega$. The molar conductivity of $0.02 \mathrm{M}$ solution of the electrolyte will be
  1. $124 \times 10^{-4} \mathrm{~S} \mathrm{~m}^2 \mathrm{~mol}^{-1}$
  2. $1240 \times 10^{-4} \mathrm{~S} \mathrm{~m}^2 \mathrm{~mol}^{-1}$
  3. $1.24 \times 10^{-4} \mathrm{~S} \mathrm{~m}^2 \mathrm{~mol}^{-1}$
  4. $12.4 \times 10^{-4} \mathrm{~S} \mathrm{~m}^2 \mathrm{~mol}^{-1}$

Solution

There is one mistake in Question paper. Assuming concentration of solution is $0.2 \mathrm{M}$ instead of $0.02 \mathrm{M}$. Since resistance of $0.2 \mathrm{M}$ is $520 \Omega$. $ \begin{aligned} \mathrm{R} & =100 \Omega \\ \mathrm{K} & =\frac{1}{\mathrm{R}}\left(\frac{\ell}{\mathrm{a}}\right) \\ 1.29 & =\frac{1}{100}\left(\frac{\ell}{\mathrm{a}}\right) \\ \left(\frac{\ell}{\mathrm{a}}\right) & =129 \mathrm{~m}^{-1} \\ \mathrm{R} & =520 \Omega, \mathrm{C}=0.2 \mathrm{M} \\ \mathrm{K} & =\frac{1}{\mathrm{R}}\left(\frac{\ell}{\mathrm{a}}\right)=\frac{1}{520}(129) \Omega^{-1} \mathrm{~m}^{-1} \\ \mu & =\mathrm{K} \times \mathrm{V}_{\text {in } \mathrm{cm}} \\ & =\frac{1}{520} \times 129 \times \frac{1000}{0.2} \times 10^{-6} \mathrm{~m}^3 \\ & =\frac{129}{520} \times \frac{1000}{0.2} \times 10^{-6} \\ & =1.24 \times 10^{-3} \\ & =12.4 \times 10^{-4} \end{aligned} $

Asked in: JEE Main 2006

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