#### Answer

(a) The average density of the mixture is $810~kg/m^3$
(b) The average density of the mixture is $840~kg/m^3$

#### Work Step by Step

(a) We can find the volume of the gasoline.
$V_g = \frac{M_g}{\rho_g}$
$V_g = \frac{0.050~kg}{680~kg/m^3}$
$V_g = 7.35\times 10^{-5}~m^3$
We can find the volume of the water.
$V_w = \frac{M_w}{\rho_w}$
$V_w = \frac{0.050~kg}{1.00\times 10^3~kg/m^3}$
$V_w = 5.0\times 10^{-5}~m^3$
We can find the average density of the mixture.
$\rho = \frac{mass}{volume}$
$\rho = \frac{M_g+M_w}{V_g+V_w}$
$\rho = \frac{0.050~kg+0.050~kg}{(7.35\times 10^{-5}~m^3)+(5.0\times 10^{-5}~m^3)}$
$\rho = 810~kg/m^3$
The average density of the mixture is $810~kg/m^3$.
(b) We can find the mass of the gasoline.
$M_g = V_g~\rho_g$
$M_g = (50\times 10^{-6}~m^3)(680~kg/m^3)$
$M_g = 0.034~kg$
We can find the mass of the water.
$M_w = V_w~\rho_w$
$M_w = (50\times 10^{-6}~m^3)(1.00\times 10^3~kg/m^3)$
$M_w = 0.050~kg$
We can find the average density of the mixture.
$\rho = \frac{mass}{volume}$
$\rho = \frac{M_g+M_w}{V_g+V_w}$
$\rho = \frac{0.034~kg+0.050~kg}{(50\times 10^{-6}~m^3)+(50\times 10^{-6}~m^3)}$
$\rho = 840~kg/m^3$
The average density of the mixture is $840~kg/m^3$.