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Solution :

Volume of one mole of ideal gas, `V_(g)` <br> `=22.4` litre `=22.4xx10^(-3) m^(3)` <br> Radius of hydrogen molecule `=1 A//2` <br> `=0.5 A=0.5xx10^(-10) m` <br> Volume of hydrogen molecule`=4//3 pi r^(3)` <br> `=4//3xx22//7(0.5xx10^(-10))^(3) m^(3)` <br> `=0.5238xx10^(-30) m^(3)` <br> One mole contains `6.023xx10^(23)` molecules. <br> Volume of one mole of hydrogen, `VH` <br> `=0.5238xx10^(-30)xx6.023xx10^(23) m^(3)=3.1548xx10^(-7) m^(3)` <br> Now `V_(g)//V_(H)=22.4xx10^(-3)//3.1548xx10^(-7)=7.1xx10^(4)` <br> The ratio is very large. This is because the interatomic separation in the gas is very large compared to the size of a hydrogen molecule.Related Video

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