Question

Vapour-liquid equilibrium of a two-component ideal solution of trichloroethene (C₂HCl₃) and trichloromethane (CHCl₃) is established at 25 °C. The mole fraction of CHCl₃ in the vapour phase is 0. 59. What is the mass fraction of C₂HCl₃ in the liquid phase? Round your answer to two significant figures.

The vapour pressures of trichloroethene and trichloromethane at 25 °C are:

Pvap,C₂HCl₃ = 73. 0 mmHg
Pvap,CHCl₃ = 199. 1 mm Hg

Answer 1

To find the mass fraction of C₂HCl₃ in the liquid phase, we can use Raoult's law. According to Raoult's law, the vapor pressure of a component in a mixture is equal to the product of its mole fraction in the mixture and its vapor pressure in the pure state. Substituting the given values into the equation allows us to calculate the mass fraction of C₂HCl₃ in the liquid phase.

Step-by-step explanation:

To find the mass fraction of C₂HCl₃ in the liquid phase, we can use Raoult's law. According to Raoult's law, the vapor pressure of a component in a mixture is equal to the product of its mole fraction in the mixture and its vapor pressure in the pure state. Mathematically, it can be expressed as P₁ = X₁ * P₁º, where P₁ is the vapor pressure of component 1 in the mixture, X₁ is its mole fraction, and P₁º is its vapor pressure in the pure state.

In this case, we know the mole fraction of CHCl₃ in the vapor phase is 0.59. Let's assume the mole fraction of C₂HCl₃ in the liquid phase is x. Using Raoult's law, we can write the equation as 0.59 * Pvap,CHCl₃ = x * Pvap,C₂HCl₃. We are given the vapor pressures of both components, which are Pvap,C₂HCl₃ = 73.0 mmHg and Pvap,CHCl₃ = 199.1 mmHg. Substituting these values into the equation, we can solve for x, which represents the mass fraction of C₂HCl₃ in the liquid phase.

Let's do the calculations:

• 0.59 * 199.1 mmHg = x * 73.0 mmHg
• x = (0.59 * 199.1 mmHg) / 73.0 mmHg
• x = 1.606

Rounding to two significant figures, the mass fraction of C₂HCl₃ in the liquid phase is approximately 1.6.