Question

A gas mixture is made by combining 6.7 g each of Ar, Ne, and an unknown diatomic gas. At STP, the mixture occupies a volume of 15.88 L. What is the molar mass of the unknown gas? molar mass: g/mol Identify the unknown gas. ON OOOO

Answer 1

Atomic weights of AY = 39.95 Ne = 2018 No.ols moles = weight No of moles of G.79 Ay'= 39.95g/mol Atomic Weight = 0.167 ma No. of moles of Ne = 6.79 = 0.332 mor 20.18 g/mol No. of mow of unknown gas =- . 6.79 Total no.06 moles de gaseas mixture = (0.167 +0.332+ 607) mor B = (0.499 + 6-71) mol 1 moke of gas occupies 22-4 2 ? more of gas occupies 15-882 = 1 md x 15.88L - - 0.709 mot 22.4L 0.709 mal = (0.499+ 6+ 7 ) mot G ? 0.21 = 31.9 [M= 32] → Oz gas PA= 0.18 atm a Pe=0.662 atm J P- P= (0.18 +0.662) atm = 0.842 atm V= 7.22 RE 0.0821 L-atra K mot' T = (49+233) K = 322 K PV = nRT - n - = PV A+B RT 0.842 atm x 7.22 Ccs21 Latm tmót x 322k n=0.229 mol A+B nc = 0.23 mal n=noethe = (0+229 +0.23) mos n = 0.459 mot p = , + Þz+ Pz - 0.18 atm +0.662 atm +3:23) * 0.842 ating p = 1.264 atm