Define relative atomic mass. A student reacts 0.245 g of a metal carbonate X₂CO₃ with 23.6 cm³ of 0.15 M hydrochloric acid. Calculate the relative atomic mass of the metal X.

The reaction between a metal carbonate X₂CO₃ and hydrochloric acid (HCl) produces a metal chloride (XCl), water, and carbon dioxide.

$$\mathrm{X_2CO_3(s) + 2HCl(aq) \rightarrow 2XCl(aq) + H_2O(l) + CO_2(g)}$$

Given: Volume of HCl = $23.6 \text{ cm}^3 = 0.0236 \text{ dm}^3$
Concentration of HCl = $0.15 \text{ M}$

$$\text{Moles of HCl} = \text{Concentration} \times \text{Volume}$$ $$\text{Moles of HCl} = 0.15 \text{ mol/dm}^3 \times 0.0236 \text{ dm}^3$$ $$\text{Moles of HCl} = \boxed{0.00354 \text{ mol}}$$

From the balanced equation, the mole ratio of X₂CO₃ to HCl is $1:2$.

\text{Moles of X_2CO_3} = \frac{1}{2} \times \text{Moles of HCl} \text{Moles of X_2CO_3} = \frac{1}{2} \times 0.00354 \text{ mol} \text{Moles of X_2CO_3} = \boxed{0.00177 \text{ mol}}

Given: Mass of X₂CO₃ = $0.245 \text{ g}$
Moles of X₂CO₃ = $0.00177 \text{ mol}$

\text{Relative molecular mass (M_r)} = \frac{\text{Mass}}{\text{Moles}} \text{M_r of X_2CO_3} = \frac{0.245 \text{ g}}{0.00177 \text{ mol}} \text{M_r of X_2CO_3} \approx \boxed{138.42 \text{ g/mol}}

We know that Mᵣ(X₂CO₃) = $2 \times \text{RAM(X)} + \text{RAM(C)} + 3 \times \text{RAM(O)}$.
Using standard atomic masses: RAM(C) = $12.01 \text{ g/mol}$ and RAM(O) = $16.00 \text{ g/mol}$.

$$138.42 = 2 \times \text{RAM(X)} + 12.01 + (3 \times 16.00)$$ $$138.42 = 2 \times \text{RAM(X)} + 12.01 + 48.00$$ $$138.42 = 2 \times \text{RAM(X)} + 60.01$$ $$2 \times \text{RAM(X)} = 138.42 - 60.01$$ $$2 \times \text{RAM(X)} = 78.41$$ $$\text{RAM(X)} = \frac{78.41}{2}$$ $$\text{RAM(X)} = \boxed{39.21 \text{ g/mol}}$$