Write the balanced chemical equation for the reaction between NaOH and HCl, and calculate the molarity of the dilute HCl solution using the provided titration data.

Molarity of $\text{NaOH}$ ($M_b$) = $0.1 \text{ M}$
Volume of $\text{NaOH}$ ($V_b$) = $25 \text{ cm}^3$ (pipetted into the conical flask)
From the table, the accurate titre readings are $27.1 \text{ cm}^3$ and $27.4 \text{ cm}^3$.
Mean titre volume of $\text{HCl}$ ($V_a$) = $\frac{27.1 \text{ cm}^3 + 27.4 \text{ cm}^3}{2} = \frac{54.5 \text{ cm}^3}{2} = 27.25 \text{ cm}^3$.
From the balanced equation, the mole ratio of $\text{NaOH}$ to $\text{HCl}$ is $1:1$. So, $n_b = 1$ and $n_a = 1$.

The formula for titration calculations is:

Where $M_a$ and $V_a$ are the molarity and volume of the acid, $M_b$ and $V_b$ are the molarity and volume of the base, and $n_a$ and $n_b$ are their respective stoichiometric coefficients from the balanced equation.

Rounding to three significant figures:

Atomic mass of H = $1.008 \text{ g/mol}$
Atomic mass of Cl = $35.45 \text{ g/mol}$
Molar mass of $\text{HCl}$ = $1.008 + 35.45 = 36.458 \text{ g/mol}$

Concentration ($\text{gdm}^{-3}$) = Molarity ($\text{mol/dm}^3$) $\times$ Molar mass ($\text{g/mol}$)
Using the unrounded molarity:
Concentration = $0.091743 \text{ mol/dm}^3 \times 36.458 \text{ g/mol}$
Concentration $\approx 3.344 \text{ g/dm}^3$

Rounding to three significant figures: