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Q.
The weight of potassium dichromate (molecular weight $= 294$) required to prepare $0.04\, N$ of $250\, mL$ solution is
TS EAMCET 2020
Solution:
$K _{2} Cr _{2} O _{7}$ mostly react in acidic medium as oxidising agent.
The reaction occurs as follows :
$Cr _{2} O _{7}{ }^{2-}+7 H ^{+}+6 e^{-} \rightarrow 2 Cr ^{3+}+7 H _{2} O$
$\because$ Reaction involve $6 e^{-}$electrons.
Thus,
Equivalent weight $=\frac{\text { Molar mass }}{\text { Number of electrons }}=\frac{294}{6}$
$=49$
$\because$ We have to prepare $0.04\, N$ of $250\, mL$ solution.
$(\because 250 \times 4=1000\, mL )$
Thus, $\frac{0.04}{4}=0.01$
and weight of $K _{2} Cr _{2} O _{7}$ is $=49 \times 0.01=0.49\, g$