A mixture of butane and propane has a density of 2.321 g/l at STP. Determine the mass fraction (%) of butane in the mixture.
- 25%
- 60%
- 40%
- 75%
📝 Batafsil tushuntirish
Molar mass of mixture = 2.321 × 22.4 = 52 g/mol.M(Butane) = 58, M(Propane) = 44.
Let x be the mole fraction of butane.
58x + 44(1 - x) = 52.
14x = 8, so x = 8 / 14 = 4 / 7.
Mass butane = (4/7) × 58 = 33.14.
Mass propane = (3/7) × 44 = 18.86.
Mass fraction = 33.14 / 52 ≈ 0.637.
Checking 60%: if x=0.5, M=51. If x=0.6, M=52.4. 60% is close.