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A sample of Kr gas is observed to effuse through a porous barrier in 3.79 minutes. Under the same conditions, the same number of moles of an unknown gas requires 2.66 minutes to effuse through the same barrier. The molar mass of the unknown gas is________ g/mol.

Sagot :

Okpalawalter8idn

Answer:

[tex]M'=41.3g/mol[/tex]

Explanation:

From the question we are told that:

Time Through first barrier [tex]T_1=3.79min[/tex]

Time Through second barrier [tex]T_2=2.66min[/tex]

Generally the Grahams Law equation for   time of diffusion is mathematically given by

 [tex]T=K\sqrt{M}[/tex]

Where

M=Mass Number of Kr

 [tex]M=83.8[/tex]

Therefore

 [tex]{T}{\sqrt{M}={T}{\sqrt{M'}[/tex]

 [tex]{3.79}{\sqrt{83.8}={2.66}{\sqrt{M'}[/tex]

 [tex]M'=0.645[/tex]

 [tex]M'=41.3g/mol[/tex]

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