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. $150\left(\right.in$ $Scm^{2}eq^{- 1}\left.$
• B. $75\left(\right.in$ $Scm^{2}eq^{- 1}\left.$
• C. $50\left(\right.in$ $Scm^{2}eq^{- 1}\left.$
• 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
$\left(\land \right)_{e q}=\left(\frac{1}{1 \times 3}\right)\times 150=50$