Assuming the density of water to be 1 g/ml, calculate the volume occupied by one molecule of water.

The correct option is A $2.98\times {10}^{-23}c{m}^3$

The density of water is approximately 1g/ml.

Thus, 1g of water occupy volume = 1 ml

The molar mass of water = 18 g/mol

​Number of moles of water in 1g = $\frac{1}{18}$ moles = 0.055 moles H${}_2$O

​Now 18 g water contain = 6.022 $\times$ 10${}^{23}$ moleculesH${}_2$O

Thus 1 g water will have $=\frac{6.022\times {10}^{23}}{18}$ molecules H${}_2$O = 0.334 $\times$ 10${}^{23}$ molecules H${}_2$O

This shows that

0.334 $\times$10${}^{23}$ molecules H${}_2$O will occupy volume = 1 ml

Therefore volume occupied by 1 H${}_2$O molecule $=\frac{1}{0.334\times {10}^{23}}=2.99$​ $\times$ 10${}^{-23}$ ml

Thus volume occupied by 1 H${}_2$O mlecule is ​2.99 ​ $\times$ 10${}^{-23}$ ml.

Hence, the correct option is $\text{A}$