Some Basic Concepts of Chemistry 2 Question 13
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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 $(2001,1 \mathrm{M})$
======= ####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 $(2001,1 \mathrm{M})$
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed (a) (molecular weight) $/ 2$
(b) (molecular weight) $/ 6$
(c) (molecular weight) $/ 3$
(d) same as molecular weight
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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 $\mathrm{K}{2} \mathrm{Cr}{2} \mathrm{O}_{7}=\frac{\text { Molecular weight }}{6}$
