Electrochemistry 2 Question 14
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14. Resistance of $0.2 \mathrm{M}$ solution of an electrolyte is $50 \Omega$. The specific conductance of the solution of $0.5 \mathrm{M}$ solution of same electrolyte is $1.4 \mathrm{~S} \mathrm{~m}^{-1}$ and resistance of same solution of the same electrolyte is $280 \Omega$. The molar conductivity of $0.5 \mathrm{M}$ solution of the electrolyte in $\mathrm{Sm}^{2} \mathrm{~mol}^{-1}$ is
======= ####14. Resistance of $0.2 \mathrm{M}$ solution of an electrolyte is $50 \Omega$. The specific conductance of the solution of $0.5 \mathrm{M}$ solution of same electrolyte is $1.4 \mathrm{~S} \mathrm{~m}^{-1}$ and resistance of same solution of the same electrolyte is $280 \Omega$. The molar conductivity of $0.5 \mathrm{M}$ solution of the electrolyte in $\mathrm{Sm}^{2} \mathrm{~mol}^{-1}$ is
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed (a) $5 \times 10^{-4}$
(b) $5 \times 10^{-3}$
(c) $5 \times 10^{3}$
(d) $5 \times 10^{2}$
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Solution:
- In order to solve the problem, calculate the value of cell constant of the first solution and then use this value of cell constant to calculate the value of $k$ of second solution. Afterwards, finally calculate molar conductivity using value of $k$ and $m$.
For first solution,
$$ k=1.4 \mathrm{Sm}^{-1}, R=50 \Omega, \quad M=0.2 $$
Specific conductance $(\kappa)=\frac{1}{R} \times \frac{l}{A}$
$$ \begin{aligned}
