The mole fraction of Ne in the effusion of gas is 0.28 .
Given ,
An equimolar mixture of Ne and Xe is accidentally placed in a container that has a tiny leak .
After a short while , a very small proportion of the mixture has escaped .
We know ,
molecular mass of Neon(Ne) = 20.1797g
molecular mass of Xenon (Xe) = 131.293g
According to the Grahan's law of diffusion,
the rate of effusion of Ne and Xe and the molecular mass of Ne and Xe are related as ,
[tex]\frac{r(Ne)}{r(Xe)} =\frac{n(Xe)}{n(Ne)} =\sqrt{\frac{M(Xe)}{M(Ne)} } =\sqrt{\frac{131.293}{20.1797} } =\frac{11.458}{4.492}[/tex]
through observing this,
Let , n(Xe) = 11.458x
    n(Ne) =4.492x
Let , X be the mole fraction of Ne .
    X = 4.492 /(11.458+4.492) = 4.492 / 15.95 = 0.28
Hence , the mole fraction of Ne in the effusion of gas is 0.28 .
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