What is the actual volume occupied by water molecules present in $20 c m^{3}$ of water ?

20 c m^{3}

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10 c m^{3}

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40 c m^{3}

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24.89 d m^{3}

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Solution

The correct option is D $24.89 d m^{3}$
Density of water = 1 $g / c c$

Therefore 20 $c m^{3}$ weights 20 g of water molecules.

So number of moles of water $= \frac{20}{18} = 1.11$ moles

1 mole occupies 22.4 L.

Therefore 1.11 moles occupies 24.89 L, i.e., 24.89 $d m^{3}$ .

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