A quantity of ammonium chloride was heated with 100 mL of 0.8 N NaOH solution till the reaction was complete. The excess of NaOH was neutralised with 12.5 mL of `0.75N H_(2)SO_(4)`. Calculate the quantity of ammonium chloride.

Explanation

12.5 mL of `0.75 N H_(2)SO_(4)-=12.5 mL " of" 0.75 N NaOH`
12.5 mL of 0.75 N NaOH `-= 11.72 mL " of" 0.8 N NaOH`
NaOH solution used by `NH_(4)Cl`
`=(100-11.72)mL ` of 0.8N NaOH
=88.28 mL of 0.8 N NaOH
`-=88.28 mL " of " 0.8 N NH_(4)Cl`
Mass of `NH_(4)Cl` present in 88.28 mL of `0.8N NH_(4)Cl` solution
`=(NxxExxV)/(1000)=(0.8xx53.5xx88.28)/(1000)=3.7783 g`
[Eq. mass of `NH_(4)Cl=53.5]`