Electrochemistry Result Question 14
14. Resistance of 0.2 M solution of an electrolyte is $50 \Omega$. The specific conductance of the solution of 0.5 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 M solution of the electrolyte in $\mathrm{Sm}^2 \mathrm{~mol}^{-1}$ is
(2014 Main)
(a) $5 \times 10^{-4}$
(b) $5 \times 10^{-3}$
(c) $5 \times 10^3$
(d) $5 \times 10^2$
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Answer:
Correct Answer: 14. ( a )
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} \Rightarrow \quad 1.4 \mathrm{Sm}^{-1} & =\frac{1}{50} \times \frac{l}{A} \\ \Rightarrow \quad \frac{l}{A} & =50 \times 1.4 \mathrm{~m}^{-1} \end{aligned} $
For second solution,
$ \begin{aligned} & R=280, \frac{l}{A}=50 \times 1.4 \mathrm{~m}^{-1} \\ & \kappa=\frac{1}{280} \times 1.4 \times 50=\frac{1}{4} \end{aligned} $
Now, molar conductivity
$ \begin{aligned} \lambda_{\mathrm{m}} & =\frac{\kappa}{1000 \times m}=\frac{1 / 4}{1000 \times 0.5}=\frac{1}{2000} \\ & =5 \times 10^{-4} \mathrm{~S} \mathrm{~m}^2 \mathrm{~mol}^{-1} \end{aligned} $
BETA
