A titration experiment involves reacting sodium carbonate (solution B) with sulfuric acid (solution A). Solution B has a concentration of 6.50 g/dm³. 25.00 cm³ of solution B required 22.50 cm³ of solu

The atomic masses are Na = 23 g/mol, C = 12 g/mol, O = 16 g/mol.

$$\text{M}(\text{Na}_2\text{CO}_3) = (2 \times 23) + 12 + (3 \times 16) = 46 + 12 + 48 = 106 \, \text{g/mol}$$

Solution B contains $6.50 \, \text{g}$ of $\text{Na}_2\text{CO}_3$ per $\text{dm}^3$.

$$\text{Concentration of B} = \frac{\text{Mass concentration}}{\text{Molar mass}} = \frac{6.50 \, \text{g/dm}^3}{106 \, \text{g/mol}} = 0.06132 \, \text{mol/dm}^3$$

The concentration of B is $\boxed{\text{0.0613 mol/dm}^3}$.

$$\text{Na}_2\text{CO}_3 (\text{aq}) + \text{H}_2\text{SO}_4 (\text{aq}) \rightarrow \text{Na}_2\text{SO}_4 (\text{aq}) + \text{H}_2\text{O} (\text{l}) + \text{CO}_2 (\text{g})$$

From the equation, the mole ratio of $\text{Na}_2\text{CO}_3 : \text{H}_2\text{SO}_4$ is $1:1$.

Given:
Concentration of B ($C_B$) = $0.06132 \, \text{mol/dm}^3$
Volume of B ($V_B$) = $25.00 \, \text{cm}^3 = 0.02500 \, \text{dm}^3$
Volume of A ($V_A$) = $22.40 \, \text{cm}^3 = 0.02240 \, \text{dm}^3$
Mole ratio ($n_A/n_B$) = $1/1$
The titration formula is $\frac{C_A V_A}{n_A} = \frac{C_B V_B}{n_B}$.

$$C_A = \frac{C_B V_B n_A}{V_A n_B} = \frac{0.06132 \, \text{mol/dm}^3 \times 0.02500 \, \text{dm}^3 \times 1}{0.02240 \, \text{dm}^3 \times 1} = 0.0684375 \, \text{mol/dm}^3$$

The atomic masses are H = 1 g/mol, S = 32 g/mol, O = 16 g/mol.

$$\text{M}(\text{H}_2\text{SO}_4) = (2 \times 1) + 32 + (4 \times 16) = 2 + 32 + 64 = 98 \, \text{g/mol}$$

$$\text{Concentration of A (g/dm}^3) = \text{Concentration of A (mol/dm}^3) \times \text{Molar mass of H}_2\text{SO}_4$$ $$\text{Concentration of A} = 0.0684375 \, \text{mol/dm}^3 \times 98 \, \text{g/mol} = 6.706875 \, \text{g/dm}^3$$

Rounding to three significant figures, the concentration of A is $\boxed{\text{6.71 g/dm}^3}$.