1.

The density of water is `1000 kg m^(-3)`. The density of water vapour at `100^(@) C` and 1 atmospheric pressure is `0.6 kg m^(-3)`. The volume of a molecule multiplied by the total number gives what is called, molecular volume. Estimate the ratio (or fraction) of the molecular volume to the total volume occupied by the water vapour under the above conditions of temperature and pressure.

Answer» Here,
density of water `= 1000 kg m^(-3)`
density of water vapour `= 0.6 kg m^(-3)`
For a given mass of water molecules, density is less if volume is large.
`:. ("volume of water vap our")/("volume of water")`
`=("density of water")/("density of water vapour")`
`(1000)/(0.6) = (10^3)/(6 xx 10^(-1)) = (1)/(6 xx 10^(-4))`
if density of bulk water and density of water molecule is same, then fraction of molecular volume to the total volume in liquid state `= 1`. As volume in vapour state has increased, therefore, fractional volume, i.e., molecular volume to the total volume is less by the same factor, i,e. `6 xx 10^(-4)`.


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