A neutral water molecule $\left({\mathrm{H}}_{2}0\right)$ in its vapour state has an electric dipole moment of magnitude $6.4 \times {10}^{-30}$ C-m. How far apart are the molecules' centres of positive and
negative charge?

There are 10 electrons and 10 protons in a neutral water molecule.
So, its dipole moment is
p = q (21) = 10 e (21)
Hence, length of the dipole, i.e., distance between centres of positive and negative charges is

$2\mathrm{l}=\frac{\mathrm{p}}{10\mathrm{e}}=\frac{6.4\times {10}^{-30}}{10\times 1.6\times {10}^{-19}}=4\times {10}^{-12} \mathrm{m}=4\mathrm{pm}$