Question

The mixture of $CuS$ (molar weight = $M_1$) and $C{u}_{2}S$ (molecular weight =$M_2$) is oxidised by $KMn{O}_4$ (molecular weight = $M_4$) in acidic medium, the product obtained are $C{u}^{2+}$ and $S{O}_{2+}$. The equivalent weights of $CuS,C{u}_{2}S$ and $KMn{O}_4$respectively, are :

• A. $\frac{M_1}{6},\frac{M_2}{8},\frac{M_3}{5}$
• B. $\frac{M_1}{3},\frac{M_2}{8},\frac{M_3}{5}$
• C. $\frac{M_1}{6},\frac{M_2}{7},\frac{M_3}{5}$
• D. $\frac{M_1}{6},\frac{M_2}{8},\frac{M_3}{7}$

Answer 1

The correct option is A $\frac{M_1}{6},\frac{M_2}{8},\frac{M_3}{5}$

For $CuS$, the oxidation number of $S$ changes from -2 to +4.
Total change in the oxidation number is +6.
Equivalent weight is the ratio of molar mass to the change in the oxidation number.

$\therefore$ The equivalent weight of $CuS$ is $\frac{M_1}{6}$

For $C{u}_{2}S$ the oxidation number of S changes from -2 to +4.
The oxidation number of Cu changes from +1 to +2.
Total change in the oxidation number is $+6+2(+1)=+8$ .

$\therefore$ The equivalent weight of $C{u}_{2}S$ is $\frac{M_2}{8}$

For $KMn{O}_4$ , the oxidation number of Mn changes from +7 to +2.
The change in the oxidation number is +5.

$\therefore$ The equivalent weight of $KMn{O}_4$ is $\frac{M_3}{5}$