$\quad 0.31 \mathrm{~g}$ of an organic dibasic acid dissolved in an alcoholic solution required 62 m

Question

$\quad 0.31 \mathrm{~g}$ of an organic dibasic acid dissolved in an alcoholic solution required 62 mL of 0.1 M NaOH for complete neutralization. The molecular mass of the acid is\

Professor PC Thomas Classes · Answer

Let $\mathrm{M}_{2}$ be the molar mass of acid. Then Number of moles of acid $=\frac{0.31}{M_{2}}$\ Number of moles of NaOH used $=$ volume in litre $	imes$ molarity $=\frac{62 	imes 0.1}{1000}=0.0062$.\ Since the acid is dibasic, 1 mole acid requires 2 moles of NaOH for neutralisation. Hence we can write, $\frac{0.31}{M_{2}}=\frac{0.0062}{2} \quad$ ie; $\mathrm{M}_{2}=100$\