Calculate the number of coulombs of positive charge in $\mathbf{250}\mathbf{}{\mathbf{cm}}^{\mathbf{3}}$of (neutral) water.

The volume of water,$\mathrm{V}=250{\mathrm{cm}}^3$.

A hydrogen atom contains one proton, and an oxygen atom contains 8 protons.

In this case, we have to find the moles of the water molecule in a volume of water, and then by using the relation between the charge, Avogadro's number and the charge value on each charge, we can find the total charge on the water.

Formulae:

The mass of a substance, $\mathrm{m}=\mathrm{\rho }\times \mathrm{V}$ (i)

The net charge of a molecule, $\mathrm{Q}=14{\mathrm{N}}_{\mathrm{A}}\mathrm{q},\text{where,q=ne}$ (ii)

The mass of the water can be calculated using the given volume of water in equation (i) as:

$\begin{array}{c}\mathrm{m}=250{\mathrm{cm}}^3\times 1.0\mathrm{g}/{\mathrm{cm}}^3\text{(}\because {\mathrm{\rho }}_{\mathrm{water}}=1.0\mathrm{g}/{\mathrm{cm}}^3\text{)}\\ \text{=250g}\end{array}$

Now, the number of the moles is

$\begin{array}{c}\mathrm{n}=250/18\text{(}\because \mathrm{molar}\mathrm{mass}\mathrm{of}\mathrm{water}\mathrm{is}18\text{)}\\ =14\mathrm{moles}\end{array}$

There are ten protons (each with charge $\mathrm{q}=+\mathrm{e}$) in each molecule of ${\mathrm{H}}_2\mathrm{O}$. So, the net charge or the total number of coulombs in the volume of water is found using equation (ii) to be

$\begin{array}{c}\mathrm{Q}=14(6.02\times {10}^{23})\left(10\right)(1.60\times {10}^{-19}\mathrm{C})\\ =1.3\times {10}^7\mathrm{C}\end{array}$

Hence, the number of coulombs present is $1.3\times {10}^7$.