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Question: One mole of \({N_2}\) gas at \(0.8atm\) takes 38 seconds to diffuse through a pinhole , whereas one ...

One mole of N2{N_2} gas at 0.8atm0.8atm takes 38 seconds to diffuse through a pinhole , whereas one mole of an unknown compound of Xenon with FF at 1.6atm1.6atm takes 57 seconds to diffuse through the same hole . Calculate the molecules formula of the compound .
(Atomic weight X=136,F=19X = 136,F = 19 )

Explanation

Solution

The ability of a gas to mix spontaneously and to form a homogenous mixture is known as diffusion . The law which deals with this phenomenon is Graham's law .

Complete answer:
Graham's law of diffusion states that under similar conditions of temperature and pressures , the rates of diffusion or effusion of different gases are inversely proportional to the square root of their densities .
Therefore , r1r2=d1d2\dfrac{{{r_1}}}{{{r_2}}} = \sqrt {\dfrac{{{d_1}}}{{{d_2}}}}
Where , r denotes the rate of diffusion and d denotes the density of the gas .
If two gases are taken at different pressures , then greater the pressure , greater is the number of molecules hitting per unit area , greater is the rate of diffusion . In such cases , Graham's law of diffusion can be written as
t2t1=P1P2d1d2=P1P2M2M1\dfrac{{{t_2}}}{{{t_1}}} = \dfrac{{{P_1}}}{{{P_2}}}\sqrt {\dfrac{{{d_1}}}{{{d_2}}}} = \dfrac{{{P_1}}}{{{P_2}}}\sqrt {\dfrac{{{M_2}}}{{{M_1}}}}
Our aim is to find the molecular formula of the Xenon gas .
We will use the above formula for calculation
Now , in the question , it is given t1=38sec,t2=57sec{t_1} = 38\sec ,{t_2} = 57\sec , Pnitrogen=0.8atm,Pgas=1.6atm{P_{nitrogen}} = 0.8atm,{P_{gas}} = 1.6atm
(P denotes pressure )
Also , we know that molecular mass of N2{N_2} is 28g/mol28g/mol .
On substituting the values in the formula , we get ,
5738=0.81.6Mgas28\dfrac{{57}}{{38}} = \dfrac{{0.8}}{{1.6}}\sqrt {\dfrac{{{M_{gas}}}}{{28}}}
On solving , we get
Mgas=252g{M_{gas}} = 252g
Let us assume the formula of Xenon gas to be XeFnXe{F_n}
Given atomic weight of X=136,F=19X = 136,F = 19
Therefore , Mgas=136+(19×n)=252{M_{gas}} = 136 + (19 \times n) = 252
n=6.16\Rightarrow n = 6.1 \approx 6
Therefore the molecular formula of the compound is XeF6Xe{F_6} .

Note:
It should be noted that according to Graham's law the time taken by a gas to diffuse is inversely proportional to its Pressure and directly proportional to the square root of their molecular mass .