Two gases A and B having molecular weight 60 and 45 respectively are enclosed in a vessel. The weight of A is 0.5 g and that of B is 0.2 g. The total pressure of the mixture is 750 mm. Calculate the partial pressure (in mm) of gas A.

Explanation

No. of moles of gas A `= ((0.50 g))/((60 gmol^(-1))) = 0.0083` mol
No. of moles of gas B `= ((0.20 g))/((45 g mol^(-1))) = 0.0044`
Total no. Of moles `= (0.0083 + 0.0044) = 0.0127` mol
Total pressure of mixture `= 0.921` bar.
Partial pressure of gas `(rhoA) = ((0.0083 mol))/((0.0127 mol)) xx (0.921 "bar") = 0.602` bar
Partial pressure of gas `B (pB) = ((0.0044 mol))/((0.0127 mol)) xx (0.921 "bar") = 0.319` bar