Calculate the concentration of HCl (Solution A) in mol dm⁻³ given the information and reaction for a titration problem.

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Titration Problem:

Assumption: Since the average titre value for solution A (HCl) is not provided, I will assume an average titre volume of $\mathbf{20.00 \text{ cm}^3}$ for the calculations.

Given Information:

• Solution A: HCl, prepared by dissolving 7.3 g of HCl gas in 1 dm$^3$ of solution.
• Solution B: Na$_2$CO$_3$, contains 2.65 g in 250 cm$^3$ of solution.
• Volume of B used for titration = 20 cm$^3$.
• Reaction: $\text{Na}_2\text{CO}_3(aq) + 2\text{HCl}(aq) \rightarrow 2\text{NaCl}(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)$
• Relative atomic masses: H=1.0, C=12.0, O=16.0, Na=23.0, Cl=35.5.
• Molar volume of gas at S.T.P. = 22.4 dm$^3$.

Molar Masses:

• Molar mass of $\text{Na}_2\text{CO}_3 = (2 \times 23.0) + 12.0 + (3 \times 16.0) = 46.0 + 12.0 + 48.0 = 106.0 \text{ g/mol}$.
• Molar mass of $\text{HCl} = 1.0 + 35.5 = 36.5 \text{ g/mol}$.

Calculations:

1. (i) Concentration of B in mol dm$^{-3}$

Step 1: Calculate the number of moles of $\text{Na}_2\text{CO}_3$ in 250 cm$^3$.

$$\text{Moles of Na}_2\text{CO}_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{2.65 \text{ g}}{106.0 \text{ g/mol}} = 0.025 \text{ mol}$$

Step 2: Calculate the concentration of $\text{Na}_2\text{CO}_3$ in mol dm$^{-3}$.

$$\text{Concentration of B} = \frac{\text{Moles}}{\text{Volume in dm}^3} = \frac{0.025 \text{ mol}}{250 \text{ cm}^3 \times \frac{1 \text{ dm}^3}{1000 \text{ cm}^3}} = \frac{0.025 \text{ mol}}{0.250 \text{ dm}^3} = 0.100 \text{ mol dm}^{-3}$$

The concentration of B is $\boxed{0.100 \text{ mol dm}^{-3}}$.

1. (ii) Concentration of A in mol dm$^{-3}$

Step 1: Calculate the number of moles of $\text{Na}_2\text{CO}_3$ in the 20 cm$^3$ portion used for titration.

$$\text{Moles of Na}_2\text{CO}_3 \text{ in 20 cm}^3 = \text{Concentration} \times \text{Volume in dm}^3$$ $$= 0.100 \text{ mol dm}^{-3} \times (20 \text{ cm}^3 \times \frac{1 \text{ dm}^3}{1000 \text{ cm}^3}) = 0.100 \text{ mol dm}^{-3} \times 0.020 \text{ dm}^3 = 0.002 \text{ mol}$$

Step 2: Use the mole ratio from the balanced equation: $1 \text{ mol of Na}_2\text{CO}_3$ reacts with $2 \text{ mol of HCl}$.

$$\text{Moles of HCl} = 2 \times \text{Moles of Na}_2\text{CO}_3 = 2 \times 0.002 \text{ mol} = 0.004 \text{ mol}$$

Step 3: Calculate the concentration of A (HCl) using the assumed average titre volume of 20.00 cm$^3$.

$$\text{Concentration of A} = \frac{\text{Moles of HCl}}{\text{Volume of A in dm}^3} = \frac{0.004 \text{ mol}}{20.00 \text{ cm}^3 \times \frac{1 \text{ dm}^3}{1000 \text{ cm}^3}} = \frac{0.004 \text{ mol}}{0.020 \text{ dm}^3} = 0.200 \text{ mol dm}^{-3}$$

The concentration of A is $\boxed{0.200 \text{ mol dm}^{-3}}$.

1. (iii) Concentration of A in g dm$^{-3}$

Step 1: Convert the molar concentration of A (HCl) to mass concentration using its molar mass.

$$\text{Concentration of A in g dm}^{-3} = \text{Concentration in mol dm}^{-3} \times \text{Molar mass of HCl}$$ $$= 0.200 \text{ mol dm}^{-3} \times 36.5 \text{ g/mol} = 7.30 \text{ g dm}^{-3}$$

The concentration of A is $\boxed{7.30 \text{ g dm}^{-3}}$.

1. (iv) Volume of hydrogen chloride gas dissolved in 1 dm$^3$ of solution A at S.T.P.

Step 1: The problem states that solution A is prepared by dissolving 7.3 g of hydrogen chloride gas in 1 dm$^3$ of solution. We need to find the volume this mass of gas would occupy at S.T.P.

$$\text{Moles of HCl gas} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{7.3 \text{ g}}{36.5 \text{ g/mol}} = 0.20 \text{ mol}$$

Step 2: Use the molar volume of gas at S.T.P. (22.4 dm$^3$/mol) to find the volume.

$$\text{Volume of HCl gas} = \text{Moles} \times \text{Molar volume at S.T.P.}$$ $$$$