A student titrates 25 cm³ of a sodium hydroxide solution (Solution C) with a hydrochloric acid solution (Solution B). Solution C has a concentration of 16 g/L. The titration results are shown in the t

The atomic masses are approximately: Na = 23.0 g/mol, O = 16.0 g/mol, H = 1.0 g/mol.

$$\text{Molar mass of NaOH} = 23.0 + 16.0 + 1.0 = 40.0 \text{ g/mol}$$

The concentration of solution C is given as 16 g per litre.

$$\text{Molarity of NaOH} = \frac{\text{Mass concentration}}{\text{Molar mass}} = \frac{16 \text{ g/L}}{40.0 \text{ g/mol}} = 0.40 \text{ mol/L}$$

$$\text{HCl (aq)} + \text{NaOH (aq)} \rightarrow \text{NaCl (aq)} + \text{H}_2\text{O (l)}$$

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

To proceed with calculating the concentration of hydrochloric acid (solution B), the titration results (specifically, the volume of hydrochloric acid used) are required. This information is not visible in the provided image.

Please provide the titration table with the volumes of hydrochloric acid used. Once that is available, the calculation can be completed using the following steps:

1. Determine the average volume of HCl used from the titration results.
2. Calculate the moles of NaOH used in the titration ($25.0 \text{ cm}^3$ of $0.40 \text{ mol/L}$ solution).
3. Use the $1:1$ mole ratio to find the moles of HCl that reacted.
4. Calculate the molarity of HCl using the moles of HCl and the average volume of HCl.