Question

In the standardisation of $N a_{2} S_{2} O_{3}$ using $K_{2} C r_{2} O_{7}$ by iodometry, the equivalent weight of $K_{2} C r_{2} O_{7}$ is

• A. $\frac{\text { molecular weight }}{2}$
• B. $\frac{\text { molecular weight }}{6}$
• C. $\frac{\text { molecular weight }}{3}$
• D. same as molecular weight

Answer 1

Answer: (B)

$Cr _{2} O _{7}^{2-}+14 H ^{+}+6 e^{-} \rightarrow 2 Cr ^{3+}+7 H _{2} O 6 I ^{-} \rightarrow 3 I _{2}+6 e^{-}$
$2 Na _{2} S _{2} O _{3}+ I _{2} \rightarrow Na _{2} S _{4} O _{6}+2 NaI$
In this reaction, Eq. wt. of
$K _{2} Cr _{2} O _{7}=\frac{\text { Molecular weight }}{3 \times 2}$
$=\frac{\text { Molecular weight }}{6}$