UPSEE Solved Model Paper 3

Let the volume of excess 0.8 N NaOH after the reaction with NH4Cl be "V" mL.
V ml of 0.8 N NaOH ~ 40 mL of 0.75 N H2SO4 V × 0.8 = 40 × 0.75 V = 37.5 mL
Volume of 0.8H NaOH considered by NH4Cl = 100 - 37.5 = 62.5 mL meq of NaOH consumed by NH4Cl
= meq of NH4Cl used 0.8 × 62.5 =

w

53.5

×1000
w = 2.675 g