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?

V ml of  0.8 N NaOH    ≅40  ml   of   0.75N    H2SO4V×0.8=40×0.75V=37.5 mlVolume  of 0.8 N NaOH consumed by NH4Cl=100-37.5=62.5 mlnumber 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