Calculate the molar mass, moles, and concentration of NaHCO₃ in solution B, given mass and volume.

The atomic masses are given as H = 1.0, C = 12.0, O = 16.0, Na = 23.0.

$$\text{Molar mass of NaHCO}_3 = (23.0 \times 1) + (1.0 \times 1) + (12.0 \times 1) + (16.0 \times 3)$$ $$\text{Molar mass of NaHCO}_3 = 23.0 + 1.0 + 12.0 + 48.0 = 84.0 \text{ g/mol}$$

The mass of NaHCO₃ is 2.10 g.

$$\text{Moles of NaHCO}_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{2.10 \text{ g}}{84.0 \text{ g/mol}} = 0.025 \text{ mol}$$

The volume of solution B is 250 cm³, which is equal to $250 \text{ cm}^3 \times \frac{1 \text{ dm}^3}{1000 \text{ cm}^3} = 0.250 \text{ dm}^3$.

$$\text{Concentration of B} = \frac{\text{Moles}}{\text{Volume (dm}^3)} = \frac{0.025 \text{ mol}}{0.250 \text{ dm}^3} = \boxed{0.10 \text{ mol dm}^{-3}}$$

To calculate parts (ii), (iii), and (iv), the average volume of solution A used in the titration (from part 1(a)) is required, as well as the specific volume of solution B used for each titration (either 20.0 cm³ or 25.0 cm³). Since these values are not provided in the question, we will assume the following to demonstrate the calculations:

• Assume 25.0 cm³ of solution B was used for each titration.
• Assume the average volume of solution A used was 24.50 cm³.

From the balanced equation: $\text{NaHCO}_3(\text{aq}) + \text{HCl}(\text{aq}) \rightarrow \text{NaCl}(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})$.
The mole ratio of NaHCO₃ to HCl is 1:1.

Moles of NaHCO₃ in 25.0 cm³ of solution B:

$$\text{Moles of NaHCO}_3 = \text{Concentration of B} \times \text{Volume of B}$$ $$\text{Moles of NaHCO}_3 = 0.10 \text{ mol dm}^{-3} \times \frac{25.0}{1000} \text{ dm}^3 = 0.0025 \text{ mol}$$

Since the mole ratio is 1:1, moles of HCl reacted = 0.0025 mol.

Concentration of A (HCl):

$$\text{Concentration of A} = \frac{\text{Moles of HCl}}{\text{Average volume of A (dm}^3)}$$ $$\text{Concentration of A} = \frac{0.0025 \text{ mol}}{\frac{24.50}{1000} \text{ dm}^3} = \frac{0.0025}{0.0245} \text{ mol dm}^{-3} \approx \boxed{0.102 \text{ mol dm}^{-3}}$$

Molar mass of HCl = H + Cl = 1.0 + 35.5 = 36.5 g/mol.

$$\text{Concentration of A in g dm}^{-3} = \text{Concentration of A in mol dm}^{-3} \times \text{Molar mass of HCl}$$ $$\text{Concentration of A in g dm}^{-3} = 0.102 \text{ mol dm}^{-3} \times 36.5 \text{ g/mol} \approx \boxed{3.72 \text{ g dm}^{-3}}$$

The concentration of A in mol dm⁻³ represents the moles of HCl in 1 dm³ of solution.
Moles of HCl in 1 dm³ of solution A = 0.102 mol.
Molar volume of gas at s.t.p. = 22.4 dm³/mol.

$$\text{Volume of HCl gas} = \text{Moles of HCl} \times \text{Molar volume of gas at s.t.p.}$$ $$\text{Volume of HCl gas} = 0.102 \text{ mol} \times 22.4 \text{ dm}^3/\text{mol} \approx \boxed{2.28 \text{ dm}^3}$$