What is the density of water at a depth where pressure is $80.0 \; atm$, given that its density at the surface is $1.03 \times 10^{3} \; kg \; m^{-3}$? (Take bulk modulus of water $B = 2.2 \times 10^{9} \; Pa$ or compressibility $45.8 \times 10^{-11} \; Pa^{-1}$)

$(1.034 \times 10^3 \; kg \; m^{-3})$ Let the pressure at the surface be $p_1 = 1.013 \times 10^5 \; Pa$ and at depth $h$ be $p_2 = 80.0 \; atm = 80.0 \times 1.013 \times 10^5 \; Pa$.
Change in pressure $\Delta p = p_2 - p_1 \approx 80.0 \times 1.013 \times 10^5 \; Pa = 8.104 \times 10^6 \; Pa$.
Bulk modulus $B = -\frac{\Delta p}{\Delta V / V_1}$, so volumetric strain $\frac{\Delta V}{V_1} = \frac{\Delta p}{B}$.
Given compressibility $\frac{1}{B} = 45.8 \times 10^{-11} \; Pa^{-1}$.
$\frac{\Delta V}{V_1} = \Delta p \times \frac{1}{B} = (8.104 \times 10^6) \times (45.8 \times 10^{-11}) \approx 3.71 \times 10^{-3}$.
Since $\rho_2 = \frac{m}{V_2}$ and $\rho_1 = \frac{m}{V_1}$, we have $V_2 = V_1(1 - \frac{\Delta V}{V_1})$.
$\rho_2 = \frac{m}{V_1(1 - \Delta V / V_1)} = \frac{\rho_1}{1 - \Delta V / V_1} \approx \rho_1 (1 + \frac{\Delta V}{V_1})$.
$\rho_2 = 1.03 \times 10^3 \times (1 + 3.71 \times 10^{-3}) \approx 1.0338 \times 10^3 \; kg \; m^{-3} \approx 1.034 \times 10^3 \; kg \; m^{-3}$.