The density of water at $4^{o}C$ is $1.0\times {10}^{3}kg{m}^{-3}$. the volume occupied by one molecule of water is approximately:

The correct option is A $3.0\times {10}^{-23}mL$

Given density $=1\times {10}^{3}kg{m}^{-3}=1g{m}^{-3}$

Molar mass of water $=18$ gm/mole

No. of moles in 1 gm $=1/18=0.55$ moles

18 gm contain molecules of $H_{2}O$ $=$ $6.022\times {10}^{23}$

1 gm contain molecules of $H_{2}O$ $=$ $\frac{6.022\times {10}^{23}}{18}=0.33\times {10}^{23}$

Volume occupied by 1 molecule of water is $=\frac{1}{0.33\times {10}^{23}}=2.99\times {10}^{-23}$