The electrical conductance of 0.04M HCl solution is $1.35 \times 10^{- 2} S .$ When measured with a conductivity cell in which the cross sectioned area of the electrodes is $0.8 c m^{2}$ and separated by a distance of 1.0cm. Calculate the specific conductance, molar conductance

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answer is 422.

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Detailed Solution

Cell constant (A) = $\frac{l}{a} = \frac{1.0}{0.8} = 1.25 c m^{- 1}$

Specific conductance (k)= conductance cell constant = $1.35 \times 10^{- 2} \times 1.25 = 1.6875 \times 10^{- 2} S c m^{- 1}$

Molar conductance = $\frac{1000 c m^{3} \times 1.6875 \times 10^{- 2} S c m^{- 1}}{0.04 m o l} = 422$

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