0.80g of impure `(NH_(4))_(2) SO_(4)` was boiled with 100mL of a 0.2N NaOH solution was neutralized using 5mL of a `0.2N H_(2)SO_(4)` solution. The percentage purity of the `(NH_(4))_(2)SO_(4)` sample is:

Options

A. `82.5`
B. `72.5`
C. `66.5`
D. `17.5`

Answer

A

Explanation

Total m.eq of `NaOH` taken = 20
m. eq of `H_(2)SO_(4) =` m.eq of `NaOH` reacted
`=(5 xx 0.2)/(25) xx 250 = 10`
m. eq of `NaOH` reacted `=20 - 10 = 10`
`(NH_(4))_(2) SO_(4)+2NaOH rarr Na_(2)SO_(4) +2NH_(3) +2H_(2)O`
m-moles of `(NH_(4))_(2)SO_(4)` reacted = 5
wt. of `(NH_(4))_(2)SO_(4) rArr 5 xx 10^(-3) xx 132 = 0.66g`
Percentage of `(NH_(4))_(2)SO_(4)` in sample
`=(0.66)/(0.8) xx 100 = 82.5`