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1.80 g of impure sample of oxalate was d...

1.80 g of impure sample of oxalate was dissolved in water and the solution made to 250 mL. On titration 20 mL of this solution required 30 mL of M/50 `KMnO_4` solution . Calculated the percentage purity of the sample.

Text Solution

Verified by Experts

The balanced chemical equation is :
`2MnO_4^(-)+5C_2O_4^(2-)+16H^(+)rarr2Mn^(2+)+10CO_2+8H_2O`
Applying morality equation ,
`((M_1V_1)/n_1)_(KMnO_4)-=((M_2V_2)/n_1)_(C_2O_4^(2-))`
`1/50xx30/2-=(M_2xx20)/5`
`M_2=(30xx5)/(50xx2xx20)=0.075M`
Molecular wt. of `C_2O_4^(2-) =88`
Wt of `C_2O_4^(2-) = 0.075 xx88=6.6g`
Amount of `C_2O_4^(2-) ` in 250 mL `=(6.6)/1000xx250=1.65g`
% Purity `=(1.65)/(1.80)xx100=91.7%`
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