Find the extra pressure 2.5 km down in the sea, taking the density of water as 1.0 × 10³ kg/m³…

Step 1. Pressure. $1000\ \text{kg/m}^3 \times 9.8\ \text{m/s}^2 \times 2500\ \text{m} = 2.45 \times 10^{7}\ \text{Pa}$.

Step 2. Squeeze. $\dfrac{2.45 \times 10^{7}\ \text{Pa}}{2.2 \times 10^{9}\ \text{Pa}} = 0.0111$, that is $1.11\%$.

Step 3. Density. $\dfrac{1000\ \text{kg/m}^3}{1 - 0.0111} = 1011\ \text{kg/m}^3$.