Answer
(a) $5\times10^{-13}kg/s$
(b)$5.83\times10^{-3}kg/m^{3}$
Work Step by Step
(a) Since the mass is conserved, the mass flow rate is the same at all points, therefore, the mass flow rate of $CCl_{4}$ is $5\times10^{-13}kg/s$
(b) Let's apply Fick's law to find the $\Delta C$
$$m=\frac{DA\Delta Ct}{L}=>\Delta C=\frac{(m/t)L}{DA}$$
$\Delta C= C_{enetring}-C_{A}=>C_{A}=C_{enetring}-\Delta C$
Let's plug known values into this equation.
$C_{A}=0.01\space kg/m^{3}-\frac{(5\times10^{-13}kg/s)(5\times10^{-3}m)}{(20\times10^{-10}m^{2})(3\times10^{-4}m^{2})}=5.83\times10^{-3}kg/m^{3}$
