Question

A mixture of nitrogen and krypton gases, at a total pressure of 666 mm Hg, contains 3.31 grams of nitrogen and 6.53 grams ofkrypton. What is the partial pressure of each gas in the mixture?

P_N2 =  mm Hg
P_Kr =  mm Hg

Answer 1

No. of moles of N2 gas = weight of N2/mol.wt of N2= 3.31/28 = 0.118

No. of mole of Ar gas = weight of Ar/Atomic.wt of Ar = 6.53/40 = 0.163

Total no. of moles = 0.118+0.163=0.281

mole fraction of N2 = no. of moles of N2/total no. of moles = 0.118/0.281=0.42

mole fraction of Ar = no. of moles of Ar/total no. of moles = 0.163/0.281=0.58

PN2 = Total pressure*mole fraction of N2 = 666*0.42=279.72mm Hg

PAr = Total pressure*mole fraction of Ar = 666*0.58=386.28mm Hg