Q38 / 200 · WBJEE 2013 · Chemistry
A conductivity cell has been calibrated with
$0.01 \mathrm{M} \quad 1: 1$ electrolyte solution (specific conductance, $k=1.25 \times 10^{-3} \mathrm{S} \mathrm{cm}^{-1}$ ) in the cell and the measured resistance was $800 \Omega$ at $25^{\circ} \mathrm{C} .$ The cell constant will be
- A$1.02 \mathrm{cm}^{-1}$
- B$0.102 \mathrm{cm}^{-1}$
- Correct answerC$1.00 \mathrm{cm}^{-1}$
- D$0.5 \mathrm{cm}^{-1}$
Solution
Given, $\mathrm{K}=1.25 \times 10^{-3} \mathrm{S} \mathrm{cm}^{-1}$
$$
\rho=\frac{1}{\mathrm{x}}=\frac{1}{1.25 \times 10^{-3}} \mathrm{S}^{-1} \mathrm{cm}
$$
From
$$
\begin{aligned}
R &=\rho \frac{1}{A} \\
800 &=\frac{1}{1.25 \times 10^{-3}} \times \frac{1}{A}
\end{aligned}
$$
(where, $\frac{1}{A}=$ cell constant)
$\therefore$
$$
\begin{aligned}
\frac{1}{A} &=800 \times 1.25 \times 10^{-3} \\
&=1 \mathrm{cm}^{-1}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.