1.

75 cc of gas was collected over mercury in a tube closed a the top by a porous plug. On standing in air for some time, and when the mercury level became constant again, the volume was found to be 123 cc. What is the molecular weight of the gas ?"" (1 litre of air weighs 1.293 g at NTP)

Answer»

Solution :Molecular weight of air = weight of 1 MOLE of air
= WT. of 22.4 LITRES of air at NTP
`= 1.293 xx 22.4`
`= 28.96`
From the GIVEN question it is clear that the time during which 75 cc of the gas diffuses out and the time during which 123 cc of air diffuses in are the same.
Thus `(V_(1))/(V_(2)) = sqrt((M_(2))/(M_(1)))`
`(75)/(123) = sqrt((28.96)/(M))`
M = 78


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