Question

Equivalent conductance and molar conductance of ${Fe}_2{\left({SO}_4\right)}_3$ are related by relation:

• A. ${\wedge }_e={\wedge }_m$
• B. ${\wedge }_{eq}=\frac{{\wedge }_m}{3}$
• C. ${\wedge }_{eq}=3{\wedge }_m$
• D. ${\wedge }_{eq}=\frac{{\wedge }_m}{6}$

Answer 1

The correct option is D ${\wedge }_{eq}=\frac{{\wedge }_m}{6}$

Relation between Equivalent conductance and molar conductance is

${\lambda }_{eq}$ = $(\frac{1}{n^{+}\times z^{+}}\times {\lambda }_m)$

$n^{+}$ = Number of Cations

$z^{+}$ = Charge on cations

In ${Fe}_2{\left({SO}_4\right)}_3$ , $n^{+}$ = 2 and $z^{+}$ = 3

So, ${\lambda }_{eq}$ = $(\frac{1}{2\times 3}\times {\lambda }_m)$

${\lambda }_{eq}$ = $(\frac{1}{6}\times {\lambda }_m)$