1.

Equal amounts of two gases of molecular weight 4 and 40 are mixed. The pressure of the mixture is 1.1 atm. The partial pressure of the light gas in this mixture is

Answer»

<P>0.55 atm
0.11 atm
1 atm
0.12 atm

Solution :No. of moles of lighter gas `= (m)/(4)`
No. of moles of heavier gas `=(m)/(40)`
Total no. of moles of `(m)/(4)+(m)/(40)=(11M)/(40)`
Mole fraction of lighter gas `((m)/(4))/((11m)/(40))=(10)/(11)`
Partial PRESSURE due to lighter gas `=P_(o)xx(10)/(11)=1.1 xx (10)/(11) =1 `atm.


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