38,08=16x/28x+28y*100
1600x=1066.24x + 1066,24y => 533,76x=1066.24y => x/y = 2/1
masa molara medie = nr moli etena/nr moli amestec * masa molara etena + nr moli CO/nr moli amestec * masa molara CO
Ex: raportul din problema 2:1
masa molara = 2/3*28+1/3*28 = 28 g/mol
C2H4 si CO au mase molare egale (28 g/mol) si orice valoare ar avea raportul, masa molara medie va avea valoarea 28 g/mol