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(A) Molarity $(M)$ of $BaCl_2 = \frac{	ext{mass} 	imes 1000}{	ext{Molar mass} 	imes 	ext{Volume in } mL} = \frac{1.0 	imes 1000}{208 	imes 200}

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$241.67, \ 120.83$

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(A) Molarity $(M)$ of $BaCl_2 = \frac{	ext{mass} 	imes 1000}{	ext{Molar mass} 	imes 	ext{Volume in } mL} = \frac{1.0 	imes 1000}{208 	imes 200} = 0.02404 \ M \approx 0.024 \ M$. Normality $(N)$ of $BaCl_2 = M 	imes 	ext{valency factor} = 0.024 	imes 2 = 0.048 \ N$. Molar conductivity $(\Lambda_m)$ = $\frac{\kappa 	imes 1000}{M} = \frac{0.0058 	imes 1000}{0.024} = 241.67 \ S \ cm^2 \ mol^{-1}$. Equivalent conductivity $(\Lambda_{eq})$ = $\frac{\kappa 	imes 1000}{N} = \frac{0.0058 	imes 1000}{0.048} = 120.83 \ S \ cm^2 \ eq^{-1}$.

The specific conductance of a solution containing $1.0 \ g$ of anhydrous $BaCl_2$ in $200 \ cm^3$ of solution is $0.0058 \ S \ cm^{-1}$. Calculate the molar and equivalent conductivity of the solution. (Molar mass of $BaCl_2 = 208 \ g \ mol^{-1}$)

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