638.0 g of `CuSO_(4)` solution is titrated with excess of 0.2 M KI solution. The liberated `I_(2)` required 400 " mL of " 1.0 M `Na_(2)S_(2)O_(3)` for complete reaction. The percentage purity of `CuSO_(4)` in the sample is

Options

A. `5%`
B. `10%`
C. `15%`
D. `20%`

Answer

B

Explanation

`CuSO_(4)+KItoI_(2)toS_(2)O_(3)^(2-)`
`2Cu^(2+)+2e^(-)toCu_(2)^(o+)`
`underline(2I^(ɵ)toI_(2)+2e^(-))`
`underline(2CuSO_(4)+4KItoCu_(2)I_(2)+2K_(2)SO_(4)+I_(2))`
`I_(2)+2S_(2)O_(3)^(2-)to2I^()+S_(4)O_(6)^(2-)`
`2 mol CuSO_(4)-=4 mol KI-=1 mol I_(2)`
`-=2 mol S_(2)O_(3)^(2-)`
`mmol CuSO_(4)-=mmol S_(2)O_(3)^(2-)`
`-=400xx1mmol=0.4mol`
Weight of `CuSO_(4)-=400xx10^(-3)xx159.5g`
`% CuSO_(4)=(400xx10^(-3)xx159.5xx100)/(638)=10%`