The molar conductivity of a solution of $ ext{AgNO}_3$ at $298 ext{ K}$ (considering the given data: concen

Question

The molar conductivity of a solution of $	ext{AgNO}_3$ at $298 	ext{ K}$ (considering the given data: concentration $= 0.5 	ext{ mol/dm}^3$, electrolytic conductivity $= 5.76 	imes 10^{-3} 	ext{ S cm}^{-1}$) is:

Accepted Answer

The molar conductivity ($\Lambda_m$) is calculated using the formula:

$\Lambda_m = \frac{\kappa 	imes 1000}{C}$

Where:
$\kappa$ (electrolytic conductivity) $= 5.76 	imes 10^{-3} 	ext{ S cm}^{-1}$
$C$ (concentration/molarity) $= 0.5 	ext{ mol/dm}^3 = 0.5 	ext{ mol L}^{-1}$

Substituting the values into the formula:
$\Lambda_m = \frac{5.76 	imes 10^{-3} 	imes 1000}{0.5}$
$\Lambda_m = \frac{5.76}{0.5} = 11.52 	ext{ S cm}^2 	ext{ mol}^{-1}$.

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