A student performs a titration. They use solution K to neutralise solution L. Use the data to determine the concentration of solution L in moles per litre.

Step 1: Calculate the volume of solution K used for each trial.

• Trial I: Volume used = Final reading - Initial reading = $5.0 \, \text{cm}^3 - 0.0 \, \text{cm}^3 = 5.0 \, \text{cm}^3$
• Trial II: Volume used = Final reading - Initial reading = $10.1 \, \text{cm}^3 - 5.0 \, \text{cm}^3 = 5.1 \, \text{cm}^3$
• Trial III: Volume used = Final reading - Initial reading = $15.3 \, \text{cm}^3 - 10.1 \, \text{cm}^3 = 5.2 \, \text{cm}^3$

Step 2: Calculate the average volume of solution K used.
The concordant titres are $5.1 \, \text{cm}^3$ and $5.2 \, \text{cm}^3$. We average these values.

$$\text{Average volume of K} = \frac{5.1 \, \text{cm}^3 + 5.2 \, \text{cm}^3}{2}$$ $$\text{Average volume of K} = \frac{10.3 \, \text{cm}^3}{2}$$ $$\text{Average volume of K} = 5.15 \, \text{cm}^3$$

The average volume of solution K used is $\boxed{\text{5.15 cm}^3}$.

b) Determine the concentration of solution L in moles per litre.

Step 1: Calculate the molar mass of sodium hydroxide (NaOH).
Given atomic masses: $Na = 23$, $O = 16$, $H = 1$.

$$M_r(NaOH) = 23 + 16 + 1 = 40 \, \text{g/mol}$$

Step 2: Calculate the concentration of solution L (NaOH) in moles per litre.
Solution L contains $2 \, \text{g}$ of NaOH per litre.

$$\text{Concentration of L} = \frac{\text{Mass concentration}}{\text{Molar mass}}$$ $$\text{Concentration of L} = \frac{2 \, \text{g/L}}{40 \, \text{g/mol}}$$ $$\text{Concentration of L} = 0.05 \, \text{mol/L}$$

The concentration of solution L is $\boxed{\text{0.05 mol/L}}$.

c) Determine the number of moles of solution L that reacted with solution K.

Step 1: Convert the volume of solution L from $\text{cm}^3$ to litres.
The volume of solution L pipetted was $25 \, \text{cm}^3$.

$$\text{Volume of L} = 25 \, \text{cm}^3 = \frac{25}{1000} \, \text{L} = 0.025 \, \text{L}$$

Step 2: Calculate the number of moles of solution L.

$$\text{Moles of L} = \text{Concentration of L} \times \text{Volume of L}$$ $$$$