How much amount of $CuS{O}_4.5H_{2}O$ required for liberation of $2.54$g $I_2$ when titrated with $KI$?

The correct option is A $4.99$ gm
Molecular weight of $CuS{O}_4.5H_{2}O=249.5gmo{l}^{-1}$

The reaction for liberation of $I_2$;

$2CuS{O}_4.5H_{2}O+KI=I_2+C{u}_2{I}_2+2K_{2}S{O}_4+5H_{2}O$

$2.54g$ of $I_2=\frac{2.54}{254}=0.01$ moles of $I_2$

Now, $1$ mole of $I_2$ is liberated from $2$ moles of $CuS{O}_4.5H_{2}O$.

$\therefore 0.01$ mole of $I_2$ will be liberated from $2\times 0.01$ mole i.e. $0.02$ mole of $CuS{O}_4.5H_{2}O$.

$\therefore$ Amount of $CuS{O}_4.5H_{2}O=0.02$ mole $\times 249.5gmo{l}^{-1}=4.99gm$

Hence, option B is correct.