Answer to Question #72992 in Physical Chemistry for LIBERTY

Question

The mole fraction of He in a gaseous solution prepared from 4.0 g of He, 6.5 g of Ar, and 10.0 g of Ne is __________.
0.11
1.5
0.6
0.2

Answer 1

æ (the mole fraction) = n(He)/n(mixture);
n = m/M;
n (He) = 4/4 = 1 mol;
n (Ar) = 6.5/40 = 0.1625 mol;
n (Ne) = 10/20 = 0.5 mol;
n (mixture) = n(He) + n(Ar) + n(Ne) = 1+0,1625+0,5=1.6625 mol;
æ (He) = 1/1.6625 = 0.6 =60 %