calculate the volume occupied by one molecule of water?

Thus 1g of water occupy volume = 1 ml

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}molecules H

_{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.
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