An impure sample of `(NH_(4))_(2)SO_(4)` is supplied to a laboratory. 3.6g of the supplied `(NH_(4))_(2)SO_(4)` salt on heating with NaOH forms `NH_(3)` gas. 100 mL of `0.5(N)H_(2)SO_(4)` is required for the complete neutralisation of the liberated `NH_(3)` gas. calculate the percentage purity of the impure sample. let us assume, the impurity of the sample is insoluble in NaOH.

Explanation

`H_(2)SO_(4)(aq)+2NH_(3)(aq)to(NH_(4))_(2)SO_(4)(aq)`
According to this equation- 1000 mL 1 (N) `H_(2)SO_(4)-=17gNH_(3)`
`therefore 100 mL0.5(N)H_(2)SO_(4)-=(17xxx100xx0.5)/(1000)=0.85g" "NH_(3)`
gt `underset(132g)((NH_(4))_(2)) SO_(4)+2NaOH to underset(34g)(2NH_(3))+Na_(2)SO_(4)+2H_(2)O`
`therefore 34gNH_(3)-=132g(NH_(4))_(2)SO_(4)`
`therefore 0.85gNH_(3)-=(132)/(34)xx0.85-=3.3g(NH_(4))_(2)SO_(4)`
`therefore` Purity off the sample `(NH_(4))_(2)SO_(4)=(3.3)/(3.6)xx100=91.66%`.