A current of 1.70 A is passed through 300 mL of 0.160 solution of ZnSO₄ for 230 sec, with a current efficiency of 90%. Find the molarity of Zn²⁺ after the deposition of Zn. Assume the volume of the solution remains constant during electrolysis

$\mathrm{Current} \mathrm{passed}=3.7\mathrm{amp}(90\%\mathrm{utiliged})$

$\mathrm{Current} \mathrm{utilised}= 3.7 \mathrm{X} \frac{90}{100}$

$=3.33 \mathrm{amp}$

$\mathrm{m}=\frac{32.5 \mathrm{X} 3.33 \mathrm{X} 230}{96500}$

$=0.26\mathrm{g} \mathrm{of} \mathrm{Zn}$

$\mathrm{Before} \mathrm{electrolysis}$

$0.16=\frac{\mathrm{W}}{161} \mathrm{X} \frac{1000}{300}$

$\mathrm{W}=7.75\mathrm{g} {\mathrm{ZnSO}}_4$

$161\mathrm{g} {\mathrm{ZnSO}}_4\mathrm{\_\_\_\_}65\mathrm{g} \mathrm{of} \mathrm{Zn}$

$7.75\mathrm{g} {\mathrm{ZnSO}}_4\mathrm{\_\_\_\_\_x}$

$\mathrm{x}=\frac{7.75}{161} \mathrm{X} 65=3.13\mathrm{g} \mathrm{of} \mathrm{Zn}$

$\mathrm{Mass} \mathrm{of} {\mathrm{Zn}}^{+2}\mathrm{after} \mathrm{electrolysis} =3.13-0.26$

$=2.89$

$\mathrm{M}=\frac{2.89}{65} \mathrm{X} \frac{1000}{300}$

$=0.15\mathrm{M}$