## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

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