Explanation

(b)
$NaCl\ \longrightarrow \ {Na}^{+} + \ {Cl}^{-}\$
molten
At cathode ${Na}^{+} + \ e^{-}\ \ \longrightarrow Na$
At anode $\mathbf{\ }{2Cl}^{-}\ - \ {2e}^{-}\ \longrightarrow \ {Cl}_{2}\$
When one mole of NaCl is electrolysed, 1 mole Na and $\frac{1}{2}$ mole ${Cl}_{2}$ is obtained.
Thus, when 230 g$\left( \frac{230}{23} = 10mol \right)Na$ is formed, the moles of
$Cl_{2}$obtained will be $= \ \frac{moles\ of\ Na}{2}\$
= $\frac{10}{2}$
= 5.0 mol