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Question

Chemistry Question on Electrochemistry

A conductivity cell has a cell constant of 0.5 cm. This cell when filled with 0.01 M NaCl solution has a resistance of 384 ohms at 25∘^{\circ}C. Calculate the equivalent conductance of the given solution.

A

130.2 Ω−1cm2(g eq)−1130.2\, \Omega^{-1} cm^2 (g \,eq)^{-1}

B

137.4 Ω−1cm2(g eq)−1137.4 \, \Omega^{-1} cm^2 (g \,eq)^{-1}

C

154.6 Ω−1cm2(g eq)−1154.6 \, \Omega^{-1} cm^2 (g \,eq)^{-1}

D

169.2 Ω−1cm2(g eq)−1169.2 \, \Omega^{-1} cm^2 (g \,eq)^{-1}

Answer

130.2 Ω−1cm2(g eq)−1130.2\, \Omega^{-1} cm^2 (g \,eq)^{-1}

Explanation

Solution

Equivalent conductance, ?eq=k×1000Normality?_{eq}=\frac{k\times1000}{Normality} where, ? (Specific conductance) = C ×la\times\frac{l}{a} =1R×la=1384×0.5\frac{1}{R}\times\frac{l}{a}=\frac{1}{384}\times0.5 =1.302?10−3ohm−1cm−1= 1.302 ? 10^{-3} ohm^{-1} cm^{-1} ∴?eq=1.302×10−3×10000.01\therefore\quad?_{eq}=\frac{1.302\times10^{-3}\times1000}{0.01} = 130.2 ohm−1cm2^{-1} cm^{2} (g eq)−1^{-1}