Diborane has the formula B2H6.
Assume that boron consists of two isotopes, containing 20% B10
5 atoms and 80% B11
5 atoms,
and that all hydrogen atoms are H1.
Molecules of diborane will therefore have relative masses of 26, 27 or 28.
In what relative proportion will molecules of diborane with masses of 26, 27 and 28 occur?

Answer:

26) E

Ratio of the two isotopes = 20% 80% = 1:4
Probability of both boron being B-10 (having mass 26) = 0.2 x 0.2 = 0.04
Probability of 1 boron being B-10 and other being B-11 (having mass 27) = (0.2 x 0.8) + (0.2 x 0.8) = 0.32
(remember there are two ways in which it can be a mixture of B-10 and B-11)
Probability of both boron being B-11 (having mass 28) = 0.8 x 0.8 = 0.64
Ratio 0.04 0.32 0.64 = 1:8:16