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

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18 c m^{- 3}

22400 c m^{- 3}

6.023 \times 1 0^{- 23} c m^{- 3}

2.9 \times 1 0^{- 23} c m^{- 3}

answer is D.

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Detailed Solution

If the density of water is

1 g / c m^{- 3}

, then the volume occupied by one molecule approximately

2.9 \times 1 0^{- 23} c m^{- 3}

.
Density of water

=

1 g / c m^{- 3}

Molar mass of water

=

18 g / mol

The Molar volume is directly proportional to molar mass and inversely proportional to density. The formula of the molar volume is expressed as
Molar volume (Vm)

= \frac{Molar mass}{Density}

\Rightarrow \frac{1818 g / mol}{1 g / c m^{- 3}}

\Rightarrow 18 c m^{3} / mol

.
Volume of 1 water molecule is equal to

Vm / N_{A} .

N_{A}

is the Avogadro number.
Water molecules have a volume of

18 / (6.022 \times 1 0^{- 23})

Volume of a single water molecule

= 2.9 \times 1 0^{- 23} c m^{- 3}

\Rightarrow

= 3 \times 1 0^{- 23} c m^{- 3}

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