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Question

A 50 mL 1.92% w/v solution of a metal ion Mn+ at. wt =60 was treated with 5.32 g hydrazinehydrate (N2H4.H2O) 90% pure and the mixture was saturated with CO2 gas when entire metal gets pecipitated as a complex [M(NH2NHCOO)n] the complex was filtered of and the filterate was titrated with M/10 KIO3 in the presence of conc. HCl according to the following equation :
N2H4+IO3+2H++ClICl+3H2O+N2
The volume of M/10 KIO3 solution needed for end point to arrive was 480 mL. Find the value of n.

A
n=1
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B
n=2
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C
n=3
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D
n=4
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Solution

The correct option is C n=3
nMn+=1.922×60 mol=0.016 mol
n hydrazinehydrate=0.9×5.33250 mol=0.096 mol
neq KIO3=4×110 N×0.48 L=0.192 eq.=neq. hydrazine reacted with KIO3
n hydrazine left after reaction with Mn+=0.1924 Mol=0.048 mol
n hydrazine hydrate reacted with Mn+=(0.0960.048) mol=0.048 mol
Mn++n H2NNHCOO[M(H2NNHCOO)n]
nMn+nhydrazine=1n=0.0160.048=13
n=3

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