Question

How much $K_{2}C{r}_2{O}_7$ (molecular weight $=294.19$ ) is required to prepare one litre of $0.1N$ solution ?

• A. 9.8036 g
• B. 7.3548 g
• C. 3.6774 g
• D. 4.903 g

Answer 1

Answer: (D)

$\because$ Equivalent weight of
$K_{2}C{r}_2{O}_7=\frac{\text{Molecular weight of }{K}_{2}C{r}_2{O}_7}{\text{Oxidation number of }Cr}$
Oxidation number of $Cr$ in $K_{2}C{r}_2{O}_7$
$2\left(+1\right)+2\left(x\right)+7\left(-2\right)=0$
$2+2x-14=0$
$2x=12$
$x=6$
Equivalent weight of $K_{2}C{r}_2{O}_7=\frac{294.19}{6}$
$=49.08$
$\frac{\text{Weight of}{K}_{2}C{r}_2{O}_7}{\text{Equivalent wt.}\left(E\right)}=\frac{N\times V}{1000}$
$w=0.1\times 1\times 49.03$
$=4.903g$