If the density of water is 1 g (cm^{-3}) then the volume occ... | Chemistry | SolveeIt

Question

If the density of water is 1 g $cm^{-3}$ then the volume occupied by one molecule of water is approximately

$18 \,cm^3$

$22400$ $cm^3$

$6.02 \times 10^{-23} \,cm^3$

$3.0 \times 10^{-23} cm^3$

Answer

$3.0 \times 10^{-23} cm^3$

Explanation

Solution

Mass of $1$ molecule of water $=\frac{GMM}{N_{0}}=\frac{18}{6.02\times10^{23}} g$ $\therefore$ Volume of $1$ molecule of water $=\frac{\text{Mass}}{\text{Density}}$ $=\frac{18\times10^{-23}}{6.02\times1}=3.0\times10^{-23}$ mL

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