Question

If a gaseous mixture is made by combining 3.96 g Ar and 2.44 g Kr in an evacuated 2.50 L container at 25.0 ∘ C, what are the partial pressures of each gas, P Ar and P Kr , and what is the total pressure, P total , exerted by the gaseous mixture?

Answer 1

Moles of Ar= 3.96g 39.948 g/mol = 0.0991 nol Moles of = 0.0291 mal Kr = 2.44g 83.798 glind PV = nRT P = nRT 0.0991 P(AV) = on mal x 0.0821 Latin mal K x 1213.15+25) 2.50 L = 10.910 alin] 0.970 alin P (kr) = 0.0291 mal x 0.0821 Latin mo k' x (213.15 +25K ) 2.50 L - 10.2849 aten 0.2849 atim Total Pressure = (0.970 + 0.2849) atm d = 11. 2549 aten 549 atin I