Question

0.16 g of dibasic acid requires 25 mL decimolar NaOH solution for complete neutralisation. The molecular mass of the acid is

• A. 256
• B. 64
• C. 32
• D. 128

Answer 1

The Correct Option is D

$0. 16 \,g$ of dibasic acid is neutralised by $=25\,m\,L$ of $0.1\,M\,NaOH$ $H_{2}X+2\,NaOH=Na_{2}X+H_{2}O$ Millimoles of acid $=\frac{1}{2} \times$ millimoles of $NaOH$ $NaOH =\frac{1}{2}\times25\times0.1=1.25$ Now, millimoles $=\frac{W_{mg}}{\text{Mol.mass}}$ or Mol mass $=\frac{W_{mg}}{\text{millimoles}}$ $=\frac{0.16 \times 1000}{1.25}$ $=128$

Answer 2

Given:
Mass of dibasic acid \(H_2X\) = 0.16 g
The volume of 0.1 M NaOH solution required for neutralization = 25 mL = 0.025 L

From the reaction equation \(H_2X + 2 NaOH \rightarrow Na_2X + 2 H_2O\), we know that 1 mole of \(H_2X\) reacts with 2 moles of NaOH.

Calculate millimoles of NaOH used:
\(\text{Millimoles of NaOH} = \frac{1}{2} \times \text{millimoles of acid}\)
\(\text{Millimoles of NaOH} = \frac{1}{2} \times 25 \times 0.1\)
\(\text{Millimoles of NaOH} = 1.25\)

Calculate millimoles of dibasic acid:
The millimoles of dibasic acid are equal to the millimoles of NaOH used (since 1 mole of \(H_2X\) reacts with 2 moles of NaOH):
\(\text{Millimoles of dibasic acid} = 1.25\)

Calculate the molecular mass of the dibasic acid:
Molecular mass (Molar mass) = Mass / Millimoles
\(\text{Molecular mass} = \frac{0.16 \times 1000 \text{ mg}}{1.25}\)

\(\text{Molecular mass} = \frac{160 \text{ mg}}{1.25}\)
\(\text{Molecular mass} = 128 \text{ g/mol}\)

So, the correct option is (D): 128