Answer
$\rho=1.05\times10^3kg/m^3$.
The density at 5.4 km is about $3\%$ larger than at the surface.
Work Step by Step
Consider a certain mass of water at the surface, and the same mass at a depth of 5.4 km.
$$\rho V=\rho_0 V_0$$
Find the density at the lower depth.
$$\rho=\frac{\rho_0 V_0}{V}$$
Equation 9–7 lets us find the change in volume due to the pressure change. B is the bulk modulus of water, given in Table 9–1.
$$\rho=\frac{\rho_0 V_0}{V_0+\Delta V}=\frac{\rho_0 V_0}{V_0+(-\Delta V_0)\frac{\Delta P}{B}}=\frac{\rho_0}{1-\frac{\rho_0gh}{B}}$$
$$\rho=\frac{1025kg/m^3}{1-\frac{(1025kg/m^3)(9.80m/s^2)(5400m)}{2.0\times10^9N/m^2}}$$
$$\rho=\frac{1025kg/m^3}{0.97}=1.05\times10^3kg/m^3$$
Take a ratio.
$$\frac{\rho}{\rho_0}=\frac{1053.5}{1025}=1.028$$
The density at 5.4 km is about $3\%$ larger than at the surface.
