The total number of electrons in \(18 \mathrm{~mL}\) of water (density \(=1 \mathrm{~g} \mathrm{~mL}^{-1}\) ) is

In \(18 \mathrm{~mL}\), number of moles of
\(\mathrm{H}_{2} \mathrm{O}=\frac{\text { mass of } \mathrm{H}_{2} \mathrm{O}}{\text { molecular mass }}\)
\(=\frac{\text { density } \times \text { volume }}{\text { molecular mass }}\)
\(=\frac{1 \times 18}{18}=1 \mathrm{~mol}\)
\(\because\) Number of moles of \(\mathrm{H}_{2} \mathrm{O}\) in \(1 \mathrm{~mol}\)
\(=6.022 \times 10^{23}\)
and number of \(\mathrm{e}^{-}\)in 1 molecule of \(\mathrm{H}_{2} \mathrm{O}\)
\(=1 \times 2+8=10\)
\(\therefore\) Number of \(\mathrm{e}^{-}\)in 1 mole of \(\mathrm{H}_{2} \mathrm{O}\)
\(=6.022 \times 10^{23} \times 10\)
\(=6.022 \times 10^{24}\)