#### Answer

The mass of the liquid is 2430 kg.

#### Work Step by Step

$P = P_0 + \rho~g~h$
$P$ is the pressure
$P_0$ is the atmospheric pressure
$\rho$ is the density of the liquid
$h$ is the depth below the surface
We first find the density of the liquid.
$P = P_0 + \rho~g~h$
$\rho~g~h = P- P_0$
$\rho = \frac{P- P_0}{g~h}$
$\rho = \frac{1.3~atm- 1.0~atm}{(9.80~m/s^2)(2.0~m)}$
$\rho = \frac{(0.3~atm)(\frac{1.01\times 10^5~N/m^2}{1~atm})}{(9.80~m/s^2)(2.0~m)}$
$\rho = 1546~kg/m^3$
We then find the volume of the vat.
$V = \pi~r^2~h$
$V = \pi~(0.5~m)^2(2.0~m)$
$V = 1.57~m^3$
We then find the mass of the liquid.
$M = \rho~V$
$M = (1546~kg/m^3)(1.57~m^3)$
$M = 2430~kg$
The mass of the liquid is 2430 kg.