Estimate the total negative charge in 1 g of water. Given that Avogadro number = $6.02 \times {10}^{23}$ and molecular weight of water= 18.

One molecule of water (${\mathrm{H}}_2\mathrm{O}$) contains 2 + 8 i.e., 10 electrons. Number of molecules in 1 g of water,

$\mathrm{n}=\frac{\mathrm{Avogadro}\mathrm{number}}{\mathrm{Atomic}\mathrm{weight}}=\frac{6.02\times {10}^{23}}{18}=3.344\times {10}^{22}$

Therefore, number of electrons in l g of water,

$\mathrm{n}\text{'}=\mathrm{n}\times 10=3.344\times {10}^{22}\times 10=3.344\times {10}^{23}$

The negative charge possessed by 1 g of water,

$\mathrm{q}=\mathrm{n}\text{'}\mathrm{e}=3.344\times {10}^{23}\times 1.6\times {10}^{-19}=5.35\times {10}^4\mathrm{C}$