A student mixes 50 cm³ of 1.0 mol/dm³ HCl with 50 cm³ of 1.0 mol/dm³ NaOH. The initial temperature of both solutions is 27°C and the final temperature is 34°C. Assuming the density of the solution is

$$\text{Total volume} = \text{Volume of HCl} + \text{Volume of NaOH}$$ $$\text{Total volume} = 50 \text{ cm}^3 + 50 \text{ cm}^3 = 100 \text{ cm}^3$$

$$\text{Mass} (m) = \text{Density} \times \text{Total volume}$$ $$m = 1.0 \text{ g cm}^{-3} \times 100 \text{ cm}^3 = 100 \text{ g}$$

$$\Delta T = \text{Final temperature} - \text{Initial temperature}$$ $$\Delta T = 34^\circ\text{C} - 27^\circ\text{C} = 7^\circ\text{C}$$

Since a change of $1^\circ\text{C}$ is equal to a change of $1 \text{ K}$, $\Delta T = 7 \text{ K}$.

$$Q = 100 \text{ g} \times 4.2 \text{ J g}^{-1} \text{ K}^{-1} \times 7 \text{ K}$$ $$Q = 2940 \text{ J}$$

The heat evolved is $\boxed{\text{2940 J}}$.

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

From the equation, 1 mole of HCl reacts with 1 mole of NaOH to produce 1 mole of $\text{H}_2\text{O}$.

Volume of HCl = $50 \text{ cm}^3 = 0.050 \text{ dm}^3$
Moles of HCl = Concentration $\times$ Volume

$$\text{Moles of HCl} = 1.0 \text{ mol dm}^{-3} \times 0.050 \text{ dm}^3 = 0.050 \text{ mol}$$

Volume of NaOH = $50 \text{ cm}^3 = 0.050 \text{ dm}^3$
Moles of NaOH = Concentration $\times$ Volume

$$\text{Moles of NaOH} = 1.0 \text{ mol dm}^{-3} \times 0.050 \text{ dm}^3 = 0.050 \text{ mol}$$

Since the moles of HCl and NaOH are equal and react in a 1:1 ratio, $0.050 \text{ mol}$ of water is formed.

The heat evolved ($Q$) is $2940 \text{ J}$. Since heat is evolved, the enthalpy change is negative.

$$\Delta H_{neut} = -\frac{Q}{\text{Moles of water formed}}$$ $$\Delta H_{neut} = -\frac{2940 \text{ J}}{0.050 \text{ mol}}$$ $$\Delta H_{neut} = -58800 \text{ J mol}^{-1}$$

$$\Delta H_{neut} = -58800 \text{ J mol}^{-1} \times \frac{1 \text{ kJ}}{1000 \text{ J}}$$ $$\Delta H_{neut} = -58.8 \text{ kJ mol}^{-1}$$

The enthalpy change per mole of water formed is $\boxed{\text{-58.8 kJ mol}^{-1}}$.