Question

A current of $96.5A$ is passed for $18\min$ between nickel electrodes in $500mL$ solution of $2MNi{\left(N{O}_3\right)}_2$. The molarity of solution after electrolysis would be

• A. 0.46 M
• B. 0.92 M
• C. 0.625 M
• D. 1.25 M

Answer 1

Answer: (B)

Moles of $Ni{\left(N{O}_3\right)}_2$ in $500mL$ of $2MNi{\left(N{O}_3\right)}_2$
is
$=\frac{2\times 500}{1000}=1mol$
Charge $=96.5\times 18\times 60=104220C$
$N{i}^{2+}+2e^{-}⟶Ni$
$2\times 96500C$ deposits $1$ mole of $Ni{\left(N{O}_3\right)}_2$
$\therefore 104220C$ will deposite
$=\frac{104220}{2\times 96500}=0.54mol$
So, moles of $Ni{\left(N{O}_3\right)}_2$ left
$=1.0-0.54=0.46mol$
Thus, molarity of $Ni{\left(N{O}_3\right)}_2$ solution
$=2\times 0.46=0.92mol/L$