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Q.
How much $ K_2Cr_2O_7 $ (molecular weight $ = 294.19 $ ) is required to prepare one litre of $ 0.1\, N $ solution ?
AMUAMU 2019
Solution:
$\because$ Equivalent weight of
$K_{2}Cr_{2}O_{7}= \frac{\text{Molecular weight of } K_{2}Cr_{2}O_{7}}{\text{Oxidation number of } Cr}$
Oxidation number of $Cr$ in $K_{2}Cr_{2}O_{7}$
$2\left(+1\right)+2\left(x\right)+7\left(-2\right)=0$
$2+2x-14=0$
$2x=12$
$x=6$
Equivalent weight of $K_{2}Cr_{2}O_{7}=\frac{294.19}{6}$
$=49.08$
$\frac{\text{Weight of} K_{2}Cr_{2}O_{7}}{\text{Equivalent wt.} \left(E\right)}=\frac{N\times V}{1000}$
$w=0.1\times1\times49.03$
$=4.903\,g$