22.6 g of an ammonium salt were treated with 100 mL of normal NaOH solution and boiled till no more of ammonia gas was given off. The excess of NaOH solution left over required 60 mL normal sulphuric acid. Calculate the percentage of ammonia in the salt.

Explanation

60 mL normal `H_(2)SO_(4)-=60 mL` normal NaOH
Thus, (100-60) mL normal NaOH were consumed by ammonium salt.
So, 40 mL normal `NaOH -= 40 mL` normal `NH_(3)`
Amount of `NH_(3)` in 40 mL normal `NH_(3)`
`=("Eq. mass of " NH_(3)xx40)/(1000)`
`=(17xx40)/(1000)=0.68`
SO, % of ammonia in the ammonium salt `=(0.68)/(2.26)xx100`
=30.09