Resistance of a conductivity cell filled with $0.1 \mathrm{~mol} \mathrm{~L}^{-1}$ $\mathrm{NaCl}$ is $100…

Resistance of a conductivity cell filled with $0.1 \mathrm{~mol} \mathrm{~L}^{-1}$ $\mathrm{NaCl}$ is $100 \mathrm{ohm}$. If the resistance of the same cell when filled with $0.02 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{NaCl}$ solution is 258 ohm, the conductivity of $0.02 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{NaCl}$ solution is (Conductivity of $0.1 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{NaCl}$ is $1.29 \mathrm{~S} \mathrm{~m}^{-1}$ )
  1. $1.0 \mathrm{~S} \mathrm{~m}^{-1}$
  2. $0.2 \mathrm{~S} \mathrm{~m}^{-1}$
  3. $2.0 \mathrm{~S} \mathrm{~m}^{-1}$
  4. $0.5 \mathrm{~S} \mathrm{~m}^{-1}$

Solution

$\begin{aligned} & \mathrm{R}_1=100 \Omega ; \mathrm{R}_1=\rho_1 \frac{\ell}{\mathrm{A}} ; \therefore \frac{\ell}{\mathrm{A}}=\mathrm{R}_1 \lambda_1=1.29 \times 100 \mathrm{~cm}^{-1} \\ & \mathrm{R}_2=258 \Omega ; \text { Now, } \frac{\ell}{\mathrm{A}}=\mathrm{R}_2 \lambda_2 ; \lambda_2=\frac{100 \times 1.29}{258}=\frac{1}{2} \mathrm{Scm}^{-1}\end{aligned}$ $=0.5 \mathrm{Scm}^{-1}$

Asked in: AP EAMCET 2022 (05 Jul Shift 2)

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