In which one of the following one Faraday of electricity liberates 1/2 gram atom of the metal ?

It is provided in the question that,
$1\mathrm{F}=\frac{1}{2}$of weight of atom
for the metal ${\mathrm{CuSO}}_4$ it dissociates
${\mathrm{CuSO}}_4\to {\mathrm{Cu}}^{+2}+2{\mathrm{e}}^{-}\to \mathrm{Cu}$
that means 2F=Cu weight of atom is 63.5
for 

$1\mathrm{F}=\frac{1}{2}\times \mathrm{weight} \mathrm{of} \mathrm{atom}=\frac{63.5}{2}$

Hence, option C is correct answer