A solution of Sodium Hydroxide containing 1 litre of 1 mol/L concentration is used to neutralize 0.245 g of a dibasic acid H2A. Calculate the relative molecular mass of the acid.

A dibasic acid ($\text{H}_2\text{A}$) reacts with Sodium Hydroxide ($\text{NaOH}$).

$$\text{H}_2\text{A} + 2\text{NaOH} \rightarrow \text{Na}_2\text{A} + 2\text{H}_2\text{O}$$

From the equation, $1 \text{ mole}$ of dibasic acid reacts with $2 \text{ moles}$ of Sodium Hydroxide.

Given:
Volume of $\text{NaOH} = 25 \text{ cm}^3 = 0.025 \text{ L}$
Concentration of $\text{NaOH} = 1 \text{ mol/L}$ (based on the interpretation of "containing 1 litre")

Moles of $\text{NaOH} = \text{Concentration} \times \text{Volume}$

$$\text{Moles of NaOH} = 1 \text{ mol/L} \times 0.025 \text{ L} = 0.025 \text{ mol}$$

Using the mole ratio from the balanced equation ($1 \text{ mole H}_2\text{A} : 2 \text{ moles NaOH}$):

$$\text{Moles of H}_2\text{A} = \frac{\text{Moles of NaOH}}{2} = \frac{0.025 \text{ mol}}{2} = 0.0125 \text{ mol}$$

Given:
Mass of dibasic acid = $0.245 \text{ g}$
Moles of dibasic acid = $0.0125 \text{ mol}$

Relative molecular mass (Molar Mass) = $\frac{\text{Mass}}{\text{Moles}}$

$$\text{Relative molecular mass of H}_2\text{A} = \frac{0.245 \text{ g}}{0.0125 \text{ mol}} = 19.6 \text{ g/mol}$$

The relative molecular mass of the acid is $\text{19.6}$.