$\begin{array}{1 1}(a)\;2.3gm\\(b)\;1.92gm\\(c)\;1.26gm\\(d)\;3.26gm\end{array}$

For calculating the minimum weight of NaOH ,which is supposed to be added to cathode compartment ,we are required to know the $[H^+]$ present in this compartment.

There occuring in the cell is

$Zn+2H^+\rightarrow Zn^{+2}+H_2(g)$

$E_{cell}=E^0_{H^+/H_2}-E^0_{Zn^{+2}/Zn}-\large\frac{0.059}{2}$$\log \large\frac{P_{H_2}[Zn^+2]}{[H^+]^2}$

$0.701=0.76-\large\frac{0.059}{2}$$\log \large\frac{1\times 0.1}{[H^+]^2}$

$\therefore [H^+]=0.0316$mol/lit

Since sufficient NaOH is to be added to cathode compartment to consume entire $H^+$,the equivalent of HCl must be equal to eq of NaOH.Let the weight of NaOH added be $x$gm.

$\large\frac{x}{40}$$=0.0316$

So $x=1.264$gm

Hence (c) is the correct answer.

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