Number of moles of K₂Cr₂O₇ reduced by one mole of Sn²⁺ ions is

$$\begin{gathered} C{r}_2{O}_7^{-2} +14H^{+} + +3S{n^{2+}}^{ } , \\ Thus 3 mole of Sn{2+ }^{ } will reduce, \\ Therefore, 1 mole of S{n^{2+}}^{ } will reduce 1/3 moles of {{}^{ }dichromate ion.}^{ } \end{gathered}$$