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

18 c m^{3}

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22400 c m^{3}

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6.03 \times 10^{- 23} c m^{3}

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3.0 \times 10^{- 23} c m^{3}

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Solution

The correct option is B $3.0 \times 10^{- 23} c m^{3}$
$d e n s i t y = \frac{m a s s}{v o l u m e}$
$v o l u m e = \frac{m a s s}{d e n s i t y}$
$= \frac{1 g r a m}{1 g r a m c m ⁻ ³} = 1 c m ³$
$1 g w a t e r = \frac{1}{18 m o l e s o f w a t e r}$
volume occupied by 1 molecules of water = $\frac{18}{(6.02 \times 10 ² ³) c m^{3}}$
$= 3 \times 10 ⁻ ² ³ c m ³$
conclusion: hence the option (D) is correct.

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