If the resistance of $0.1 \mathrm{M} \mathrm{KCl}$ solution in a conductance cell is $300 \Omega$ and…

If the resistance of $0.1 \mathrm{M} \mathrm{KCl}$ solution in a conductance cell is $300 \Omega$ and conductivity is $0.013 \mathrm{Scm}^{-1}$, then the value of cell constant is
  1. $3.9 \mathrm{~cm}^{-1}$
  2. $39 \mathrm{~m}^{-1}$
  3. $3.9 \mathrm{~m}^{-1}$
  4. $0.39 \mathrm{~cm}^{-1}$

Solution

Conductivity $(\kappa)=\frac{1}{\text { Resistance }(R)} \times$ cell constant Cell constant $=\kappa \times R$ Given, $\kappa=0.0135 \mathrm{~cm}^{-1} \Rightarrow R=300 \Omega$ Cell constant $=0.0135 \times 300=3.9 \mathrm{~cm}^{-1}$. .

Asked in: JEE-TOPICTESTS-CHEMISTRY

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