Study Companion

For more NEET and AIIMS mcq on Chemical arithmetic and volumetric analysis Visit
Q73) 1.25g of solid dibasic acid is completely neutralised by 25mL of 0.25 molar Ba(OH)2 solution. Molecular mass of the acid is ... [ NTSE 1991]


0.25 M Ba(OH)2=0.5N Na(OH)2 [ As it's acidity is 2]
25 mL of 0.5 N Ba(OH)2 neutralize acid=1.25 g
∴ 25 mL of 1 N Ba(OH)2 will neutralize acid=2.5 g
∴ 1000 mL of 1 N Ba(OH)2 will neutralize acid=100 g
Calculation is on the basis of 1 N solution, Eq. Wt of acid=100
and Molecular weight=100 × 2=200
Answer: (d)