Question

Calculate the volume occupied by one molecule of water? (Given density 1 g cm^-3)

Answer 1

Step: 1 Calculating the volume occupied by one mole of water:

Given to us that Density of water: ${\mathrm{d}}_{\mathrm{w}}=1\mathrm{g}{\mathrm{cm}}^{-3}$

$\mathrm{Density}=\frac{\mathrm{Mass}}{\mathrm{Voume}}$

$$\begin{gathered} \mathrm{Volume}\mathrm{of}\mathrm{water}=18\times 1{\mathrm{cm}}^3/\mathrm{mol} \\ \mathrm{Volume}\mathrm{of}\mathrm{water}=18{\mathrm{cm}}^3/\mathrm{mol} \end{gathered}$$

Step: 2 Calculating the volume occupied by one molecule of water:

$$\begin{gathered} Volumeofone=\frac{18}{6.022\times {10}^{23}}=2.98\times {10}^{-23}c{m}^3 \\ moleculeofwater \end{gathered}$$

Therefore, the volume occupied by one molecule of water is $2.98\times {10}^{-23}c{m}^3$.