Question

Equal masses of sulphur dioxide and oxygen are mixed in an empty container at 25 degree Celsius. What is the fraction of the total pressure exerted by sulphur dioxide?

Answer 1

Molar mass of Sulphur dioxide (SO₂) = 32 + 2(16) =64 g/mol
Molar mass of O₂ = 32 g
Equal mass of SO₂ and O₂ are mixed. Let the mass be "x" g
Number of moles of SO₂ = $\frac{x}{64}moles$
Number of moles of oxygen = $\frac{x}{32}moles$
Fraction of total pressure exerted is directly proportional to the mole fraction of the component.
pH₂=partial pressure exerted by components=mole fraction $\times$total pressure
Hence the fraction of pressure exerted by the SO₂=$\frac{{\displaystyle \frac{x}{64}}}{{\displaystyle \frac{x}{64}}+{\displaystyle \frac{x}{32}}}=\frac{{\displaystyle \frac{x}{64}}}{{\displaystyle \frac{x+2x}{64}}}=\frac{1}{3}$
Fraction of total pressure exerted by oxygen = $\frac{{\displaystyle \frac{x}{32}}}{{\displaystyle \frac{x}{32}}+{\displaystyle \frac{x}{64}}}=\frac{{\displaystyle \frac{x}{32}}}{{\displaystyle \frac{2x+x}{64}}}=\frac{x\times 64}{3x\times 32}=\frac{2}{3}$