Resistance of $0.05 \ M$ electrolytic solution at $298 \ K$ temperature is $30.0 \ \Omega$. The cross-sectional area of the conductivity cell having $Pt$ electrodes is $3.8 \ cm^{2}$ and the distance between the two electrodes is $1.5 \ cm$. What is the molar conductivity of the electrolytic solution?
(N/A) Cell constant $(G^{*})$ is given by:
$G^{*} = \frac{l}{A} = \frac{1.5 \ cm}{3.8 \ cm^{2}} = 0.3947 \ cm^{-1}$
Conductivity $(k)$ is calculated as:
$k = \frac{G^{*}}{R} = \frac{0.3947 \ cm^{-1}}{30.0 \ \Omega} = 0.013157 \ S \ cm^{-1}$
Molar conductivity $(\Lambda_{m})$ is calculated using the formula:
$\Lambda_{m} = \frac{1000 \times k}{c}$
$\Lambda_{m} = \frac{1000 \times 0.013157 \ S \ cm^{-1}}{0.05 \ mol \ L^{-1}} = 263.14 \ S \ cm^{2} \ mol^{-1}$
$G^{*} = \frac{l}{A} = \frac{1.5 \ cm}{3.8 \ cm^{2}} = 0.3947 \ cm^{-1}$
Conductivity $(k)$ is calculated as:
$k = \frac{G^{*}}{R} = \frac{0.3947 \ cm^{-1}}{30.0 \ \Omega} = 0.013157 \ S \ cm^{-1}$
Molar conductivity $(\Lambda_{m})$ is calculated using the formula:
$\Lambda_{m} = \frac{1000 \times k}{c}$
$\Lambda_{m} = \frac{1000 \times 0.013157 \ S \ cm^{-1}}{0.05 \ mol \ L^{-1}} = 263.14 \ S \ cm^{2} \ mol^{-1}$
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