Question

One mole of nitrogen gas at ${0.8atm}$ takes ${38 seconds}$ to diffuse through a pinhole whereas one mole of an unknown compound of xenon with fluorine at ${1.6 atm}$ takes ${57 seconds}$ to diffuse through the same hole.
Calculate the molecular formula of the compound.

Answer 1

Graham's law of diffusion states that under similar conditions of temperature and pressure, the rate at which gases diffuse is inversely proportional to the square root of their densities.

Complete answer:
It is given that;
${ t }_{ 1 }$ = ${ 38 seconds }$
${ t }_{ 2 }$ = ${ 57 seconds }$
${ P }_{ 1 }$ = ${ 0.8atm }$
${ P }_{ 2 }$ = ${ 1.6atm }$
${ M }_{ 1 }$ = ${ 28gmol }^{ -1 }$
${ M }_{ 2 }$ =?

From Graham’s law of diffusion;
${ t }_{ 2 }{ \div t }_{ 1 }{ =P }_{ 1 }{ \div P }_{ 2 }\sqrt { { M }_{ 2 }{ \div M }_{ 1 } }$........(1)
${ t }_{ 1 }$ = time taken by nitrogen gas
${ t }_{ 2 }$ = time taken by the compound
${ P }_{ 1 }$ = pressure of nitrogen gas
${ P }_{ 2 }$ = pressure of the compound
${ M }_{ 1 }$ = molecular formula of nitrogen gas
${ M }_{ 2 }$ = molecular formula of the compound.

Now, putting the given values in equation (1), we get
${ 57\div 38=0.8\div 16 }\sqrt { { M }_{ 2 }{ \div 28 } }$

Now, squaring both the sides
${ (57\div 38{ ) }^{ 2 }=1\div 4 }{ \times M }_{ 2 }{ \div 28 }$
${ M }_{ 2 }{ =(57 }^{ 2 }{ \times 28\times 4\div { 38 }^{ 2 } }$
The molecular weight of the compound = ${ 252 }$
The compound contains Xe and F , so Xe +nF = ${ 252 }$
The molecular mass of Xe = ${ 131 }$
The molecular mass of F = ${ 19 }$
Hence, ${ 131+ n(19) = 252 }$
We get, n =${ 6 }$
So, the molecular formula of the compound is ${ XeF }_{ 6 }$.

Additional Information:
Graham's law is generally precise for molecular effusion which includes the movement of each gas at a time through a hole. It is just approximate for the diffusion of one gas in another or in air, as these procedures involve the movement of more than one gas.

Note: The possibility to make a mistake is that Graham’s law is equal to the square root of the molar mass of gas 2 by the molar mass of gas 1.