Question

Molar conductivity of a solution of an electrolyte $AB_{3}$ is $150$ $Scm^{2}mol^{- 1}$ . If it ionizes as $\text{AB}_{3} \rightarrow \text{A}^{3 +} + 3 \text{B}^{-}$ , its equivalent conductivity will be

• A. $150\left(\right.$ in $\left. Scm ^{2} eq ^{-1}\right)$
• B. $75\left(\right.$ in $\left. Scm ^{2} eq ^{-1}\right)$
• C. $50\left(\right.$ in $\left. Scm ^{2} eq ^{-1}\right)$
• D. $80\left(\right.$ in $Scm ^{2} eq ^{-1}$ )

Answer 1

Answer: (C)

$\land _{e q}=\left[\frac{1}{n^{+} \times Z^{+}}\right]\land _{m}$

Where, $\land _{m}=$ molar conductance

$n^{+}=$ number of cations

$Z^{+}=$ charge of cation

$\wedge_{\mathrm{eq}}=\left(\frac{1}{1 \times 3}\right) \times 150=50$