A 50ml 1.92% (w/v) solution of a metal ion $M^{n +}$ (atomic weight =60) was treated with 5.332 g hydrazine hydrate (90 % pure) and mixture was saturated with $C O_{2}$ gas when entire metal gets precipitated as a complex $[M {(H_{2} N - N H C O O)}_{n}]$ . The complex was filtered off and the filtrate was titrated with $\frac{M}{10} K I O_{3}$ in the presence of conc. HCl according to the following equation $N_{2} H_{4} + I O_{3}^{-} + 2 H^{+} + 2 C l^{-} \to I C l + 3 H_{2} O + N_{2} ↑$ the volume of $\frac{M}{10} K I O_{3}$ solution needed for the end point arrive was 480 ml. Find the value of n.

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Detailed Solution

$n_{M^{n +}} = \frac{1.93}{2 \times 60} m o l = 0.016 m o l$

$\begin{array}{l} n_{H y d r o z i n e h y d r a t e e} = \frac{0.9 \times 5.332}{50} m o l = 0.09 \\ n_{e q} K I O_{3} = 4 \times \frac{1}{10} N \times 0.48 L = 0.192 e q = n_{e q} h y d r a z i n e r e a c t e d w i t h K I O_{3} \\ ∴ n_{h y d r a z i n e} l e f t a f t e r r e a c t i o n w i t h M^{n +} = \frac{0.192}{4} m o l = 0.048 m o l \\ ∴ n_{H y d r a z i n e} h y d r a t e r e a c t e d w i t h M^{n +} = (0.096 - 0.048) = 0.0048 m o l \\ M^{n +} + n H_{2} N N H C O O^{-} \to [M {(H_{2} N N H C O O)}^{n}] \end{array}$