#### Answer

$q = -5.36\times 10^7~C$

#### Work Step by Step

We can find the mass of the water.
$m = \rho~V$
$m = (1000~kg/m^3)(1.0~L)(\frac{1~m^3}{10^3~L})$
$m = 1.0~kg$
Each water molecule has 10 protons and 8 neutrons. We can find the number $N$ of water molecules.
$N = \frac{1.0~kg}{(18)(1.66\times 10^{-27}~kg)}$
$N = 3.35\times 10^{25}$
Each water molecule has ten electrons. Each electron has a charge of $-1.6\times 10^{-19}~C$. We can find the total charge of all the electrons.
$q = (3.35\times 10^{25})(10)(-1.6\times 10^{-19}~C)$
$q = -5.36\times 10^7~C$