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

Given : 40 mol% Argon, 20 mole % C and balance has to be B= 100-(40+20)= 40%

Consider 1 mole of mixture

it contains = 40% Argon= 0.4 mol Argon , 0.2 mole C and 0.4 moles of B

Mass of Argon = Moles* Molecular weight= 0.4*40= 16gm

Mass of C= 0.2*50= 10 gm

But 18.75% is B and 100-18.75% = 81.25% is Argon and C

81.25% correspond to 16 (Argon)+ 10gm C= 26gms of mixture

18.75% B correspond to 18.75*26/81.25 =6 gms

But moles of B= 0.4

Mass/ Molecular weight= 0.4

Molecular weight of B= 6/0.4= 15

Total mass of Mixture = 6 mass of B+ 26 ( Argon)+C =32 gms

Moles= mass/ molecular weight

1= mass/ molecular weight ( Basis is 1 mole)

Moelcular weight = 32 g/mole