Answer
$$\Delta V=-0.0527m^{3}$$
$$\rho =1.09\times10^{3}Kg/m^{3}$$
Work Step by Step
$\frac{1}{K}=B=\frac{-\Delta P}{\frac{\Delta V}{V_{0}}}\quad\rightarrow \qquad\Delta V=\frac{-\Delta P V_{0}}{B}$
$B=\frac{1}{K} =\frac{1}{45.8\times10^{-11}}=2.2\times10^{9}Pa$
$V=1 m^{3}\qquad \Delta P=1.6\times10^{8}Pa\qquad K=45.8\times10^{-11}$
$a) \quad \Delta V=\frac{-(1.6\times10^{8})(1m^{3})}{(2.2\times10^{9})}=-0.0527m^{3}$
$b)\quad \rho=\frac{m}{V_{F}}=\frac{1.03\times10^{3}Kg}{(1-0.0.527m^{3})}=1.09\times10^{3}Kg/m^{3}$
