#### Answer

There are $3.35\times 10^{26}$ protons in 1.0 L of liquid water.

#### Work Step by Step

We can convert the volume to units of $m^3$:
$V = (1.0~L)(\frac{1~m^3}{10^3~L}) = 1.0\times 10^{-3}~m^3$
We then find the mass of this volume of water:
$m = \rho~V$
$m = (1000~kg/m^3)(10^{-3}~m^3)$
$m = 1.0~kg$
We then find the number $N$ of water molecules:
$N = \frac{1.0~kg}{(18)(1.66\times 10^{-27}~kg)}$
$N = 3.35\times 10^{25}~molecules$
There are 10 protons and 8 neutrons in each water molecule. We can find the total number of protons.
$(10)(3.35\times 10^{25}) = 3.35\times 10^{26}~protons$
Therefore, there are $3.35\times 10^{26}$ protons in 1.0 L of liquid water.