Answer to Question #94949 in Field Theory for abdul

Question

Calculate the number of coulombs of positive charge in 304 cm3 of (neutral) water. (Hint: A hydrogen atom contains one proton; an oxygen atom contains eight protons.)

Answer 1

First, let us calculate the number of $H_2 O$ molecules in the given volume of water. From one side, the amount of moles is the mass, divided my molar mass, and from the other side, it is the number of molecules, divided by Avogadro number: $\nu = \frac{m}{M} = \frac{\rho V}{M} = \frac{N}{N_A}$. From the last equation, the number of water molecules is $N = \frac{\rho V N_a}{M}$ . Substituting $\rho = 1000\frac{kg}{m^3}$ , $V = 304 cm^3 = 304 \cdot 10^{-6} m^3$, $N_A = 6.02 \cdot 10^{23} mol^{-1}$, $M = (8 +2) \frac{g}{mol} = 10\frac{g}{mol} = 10^{-2} \frac{kg}{mol}$, obtain $N \approx 1.83 \cdot 10^{25}$. Since each molecule has 10 protons, total charge is the number of molecules, multiplied by 10 and by the charge of proton: $10 \cdot 1.83\cdot 10^{25} \cdot 1.6 \cdot 10^{-19} = 2.93 \cdot 10^{7} C$.