Calculate the mass of ethanoic acid.

Given:
Volume of ethanoic acid = $15.0 \text{ cm}^3$
Density of ethanoic acid = $1.05 \text{ g/cm}^3$

Mass = Density $\times$ Volume

$$\text{Mass of CH}_3\text{COOH} = 1.05 \text{ g/cm}^3 \times 15.0 \text{ cm}^3 = 15.75 \text{ g}$$

Atomic masses: C = 12, H = 1, O = 16
Formula: $\text{CH}_3\text{COOH}$ (which is $\text{C}_2\text{H}_4\text{O}_2$)

$$\text{Molar mass of CH}_3\text{COOH} = (2 \times 12) + (4 \times 1) + (2 \times 16)$$ $$\text{Molar mass of CH}_3\text{COOH} = 24 + 4 + 32 = 60 \text{ g/mol}$$

Given:
Moles of ethanoic acid = $0.2625 \text{ mol}$
Total volume of solution = $500 \text{ cm}^3 = 0.500 \text{ L}$

Concentration = $\frac{\text{Moles}}{\text{Volume (L)}}$

$$\text{Concentration} = \frac{0.2625 \text{ mol}}{0.500 \text{ L}} = 0.525 \text{ mol/L}$$

The concentration of the solution is $\text{0.525 mol/L}$.