Question

The equivalent weight of $K_{2}C{r}_2{O}_7$ in acidic medium

• A. 294
• B. 49
• C. 298
• D. 50

Answer 1

Answer: (B)

$K_{2}C{r}_2{O}_7+3H_{2}S{O}_4\to K_{2}S{O}_4+C{r}_2{\left(S{O}_4\right)}_3+3(O)+3H_2$
No. of electrons lost $=12-6=6$
$\therefore$ Equivalent weight $=\frac{M}{6}=\frac{294}{6}=49.$