Question

A gas mixture consists of three components: argon (Ar), B, and C. The following analysis of this mixture is given: 40.0 mole % argon, 18.75 mass % B, 20.0 mole % C. The molecular weight of Ar is 40 and the molecular weight of C is 50. a. What is the molecular weight of B? b. What is the average molecular weight of the mixture?

Answer 1

Solution :-

Lets assume we have 1 mole total gas mixture

Then moles of Ar = 40% * 1 / 100 % = 0.40 mol Ar

Moles of C = 20 % * 1 /100 % = 0.20 mol C

Now lets calculate the mass of the Ar and C

Mass of Ar = moles *molar mass

                   = 0.4 mol * 39.948 g per mol

                   = 15.9792 g Ar

Mass of C = 0.20 mol * 50 g per mol

            = 10 g

So the total mass of the Ar + C = 15.9792 g + 10.0 g = 25.9792 g

This total is 100 % - 18.75 % = 81.25 % of the mass

So now lewts calculate the mass of the total sample

25.9792 g * 100 % / 81.25 % = 31.9744 g

Now lets find the mass of B

Mass of B = 31.9744 g – 25.9792 g = 5.9952 g

Now lets find the moles of the B

Moles of B = total moles – (moles of Ar + moles of C)

                     = 1 mol – (0.40 mol + 0.20 mol)

                    = 0.4 mol

Now lets calculate the molar mass of the B

Molar mass of B = 5.9952 g / 0.4 mol

                              = 14.988 g per mol

so molecular weight of B = 14.988 g/mol

So the total molar mass of mixture = 104.936 g per mol