Burette reading (initial and final) must be given to two decimal places. Volume of pipette used must be recorded but no account of experimental procedure is...

Titration of A (HNO₃) against B (Na₂CO₃)

The concentration of A is fixed by the data: molar mass of HNO₃ \( = 1+14+48 = 63\ \text{g mol}^{-1}\), so

\[ [A] = \frac{6.3}{63} = 0.10\ \text{mol dm}^{-3} \]

The actual average titre depends on the candidate's burette readings. Taking a representative average titre of \(V_A = 25.00\ \text{cm}^3\) of A for 20.0 cm³ of B, the working is:

Equation: \[ 2HNO_3 + Na_2CO_3 \to 2NaNO_3 + CO_2 + H_2O \]

(i) Concentration of B in mol dm^-3

Moles of HNO₃ used \( = 0.10 \times \dfrac{25.00}{1000} = 2.5 \times 10^{-3}\ \text{mol} \)

Mole ratio HNO₃ : Na₂CO₃ = 2 : 1, so moles of Na₂CO₃ \( = \dfrac{2.5\times10^{-3}}{2} = 1.25\times10^{-3}\ \text{mol} \) in 20.0 cm³.

\[ [B] = \frac{1.25\times10^{-3}}{20.0/1000} = \mathbf{0.0625\ mol\,dm^{-3}} \]

(ii) Concentration of B in g dm^-3

Molar mass of Na₂CO₃ \( = (2\times23)+12+(3\times16) = 106\ \text{g mol}^{-1} \)

\[ = 0.0625 \times 106 = \mathbf{6.63\ g\,dm^{-3}} \]

(iii) Mass of sodium ions in 1.0 dm³ of B

Each Na₂CO₃ gives 2 Na⁺, so moles of Na⁺ \( = 2 \times 0.0625 = 0.125\ \text{mol} \)

\[ \text{Mass of Na}^+ = 0.125 \times 23 = \mathbf{2.88\ g} \]

(The numerical answers scale with the candidate's actual average titre; the method is exactly as shown.)