Question

The molar mass of $CuS{O}_4.5H_{2}O$ is $249$.

Its equivalent mass in the reaction (a) and (b) would be:

(a)$CuS{O}_4$ + $KI⟶$ product

(b)Electrolysis of $CuS{O}_4$ solution

• A. (a) 249; (b) 249
• B. (a)124.5; (b) 124.5
• C. (a)249; (b) 124.5
• D. (a)124.5; (b) 249

Answer 1

The correct option is C (a)249; (b) 124.5

$2Cu{SO}_4+4KI\to 2CuI+I_2+2K_2{SO}_4$

$+2$ $+1$

$\therefore$ Change in oxidation number, $n=1$

Equivalent wt of Cu $=\frac{249}{1}=249g$

For electrolysis of ${CuSO}_4$,

${Cu}^{2+}+2e^{-}\to Cu$

$\therefore$ Change in oxidation number, $n=1$

equivalent wt $=\frac{249}{2}=124.5g$

Hence, option $\left(C\right)$ is correct.