The molar conductivity of a $0.5 ext{ mol/dm}^3$ solution of $ ext{AgNO}_3$ with electrolytic conductivity

Question

The molar conductivity of a $0.5 	ext{ mol/dm}^3$ solution of $	ext{AgNO}_3$ with electrolytic conductivity of $5.76 	imes 10^{-3} 	ext{ S cm}^{-1}$ at $298 	ext{ K}$ is:

Accepted Answer

Given:
Concentration, $C = 0.5 	ext{ mol/dm}^3 = 0.5 	ext{ mol/L}$
Conductivity, $\kappa = 5.76 	imes 10^{-3} 	ext{ S cm}^{-1}$

The formula for molar conductivity ($\Lambda_m$) is:
$\Lambda_m = \frac{\kappa 	imes 1000}{C}$

Substituting the given values:
$\Lambda_m = \frac{5.76 	imes 10^{-3} 	imes 1000}{0.5} = \frac{5.76}{0.5} = 11.52 	ext{ S cm}^2	ext{mol}^{-1} \approx 11.5 	ext{ S cm}^2/	ext{mol}$

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