Some Basic Concepts of Chemistry - Result Question 12
13. In the standardisation of $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ using $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ by iodometry, the equivalent weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ is (2001,1M)
(a) (molecular weight) $/ 2$
(b) (molecular weight)/6
(c) (molecular weight) $/ 3$
(d) same as molecular weight
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Answer:
Correct Answer: 13. (b)
Solution:
- The following reaction occur between $\mathrm{S}_2 \mathrm{O}_3^{2-}$ and $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ :
$26 \mathrm{H}^{+}+3 \mathrm{~S}_2 \mathrm{O}_3^{2-}+4 \mathrm{Cr}_2 \mathrm{O}_7^{2-} \longrightarrow 6 \mathrm{SO}_4^{2-}+8 \mathrm{Cr}^{3+}+13 \mathrm{H}_2 \mathrm{O}$
Change in oxidation number of $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ per formula unit is 6 (it is always fixed for $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ ).
Hence, equivalent weight of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7=\frac{\text { Molecular weight }}{6}$
BETA
