JEE MainChemistryModeratePhysical Chemistry

Molar conductivity of 0.04 solution at 298 k is 200 Scm2mole-1.
A conductivity cell which is filled with MgCl2 have area of
cross-section of electrode 4cm2 & distance between electrode is 8 cm. If
potential difference between electrode is 10V then find current flow in
miliampere.

Show answer and solution

\lambda_{m} = \frac{K \times 1000}{M}

200 = \frac{K \times 1000}{0.04}

K = \frac{200 \times 0.04}{1000} = 8 \times 10^{- 3}S{cm}^{- 1}

K = \left( \frac{\mathcal{l}}{9} \right) \times \frac{1}{R}

8 \times 10^{- 3} = \left\lbrack \frac{8}{4} \right\rbrack \times \frac{1}{R}

R = \frac{2}{8 \times 10^{- 3}} = \frac{1}{4} \times 10^{3}\Omega

Again

\Rightarrow V = IR

I = \frac{V}{R} = \frac{10}{\left( \frac{1}{4} \times 10^{3} \right)} = 4 \times 10^{- 2}A

Source: Grass-G / CH-23 Chemistry Paper 31 Dec. FST.docx

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