Question

The weight of potassium dichromate (molecular weight $=294$) required to prepare $0.04N$ of $250mL$ solution is

• A. 2.94 g
• B. 29.4 g
• C. 0.98 g
• D. 0.49 g

Answer 1

Answer: (D)

$K_{2}C{r}_2{O}_7$ mostly react in acidic medium as oxidising agent.
The reaction occurs as follows :
$C{r}_2{O}_7{}^{2-}+7H^{+}+6e^{-}\to 2C{r}^{3+}+7H_{2}O$
$\because$ Reaction involve $6e^{-}$electrons.
Thus,
Equivalent weight $=\frac{\text{ Molar mass }}{\text{ Number of electrons }}=\frac{294}{6}$
$=49$
$\because$ We have to prepare $0.04N$ of $250mL$ solution.
$(\because 250\times 4=1000mL)$
Thus, $\frac{0.04}{4}=0.01$
and weight of $K_{2}C{r}_2{O}_7$ is $=49\times 0.01=0.49g$