A certain quantity of Ammonium chloride is boiled with 100 ml of 0.8 N NaOH till no further action occurs. Excess of NaOH required 40 ml of 0.75 N sulphuricacid to neutralise if. How much Ammonium chloride was used?

$\mathrm{V}\mathrm{ml}\mathrm{of}0.8\mathrm{N} \mathrm{NaOH}\cong 40\mathrm{ml}\mathrm{of}0.75\mathrm{N}{\mathrm{H}}_2{\mathrm{SO}}_4$

$\mathrm{V}\times 0.8=40\times 0.75$

$V=37.5ml$

Volume  of 0.8 N NaOH consumed by ${\mathrm{NH}}_4\mathrm{Cl}$

=100-37.5

=62.5 ml

number of moles of NaOH=0.05 therefore number of moles of NH₄Cl will  be 0.05 hence 53.5 x 0.05 = 2.675 grams