The partial pressure of ethane over a solution containing 6.56 × 10–3 g of ethane is 1 bar. If the solution contains 5.00 × 10–2 g of ethane, then what shall be the partial pressure of the gas?

To determine the partial pressure of ethane over a solution containing 5.00 × 10–2 g of ethane, we can use Raoult's Law. This law states that the partial pressure of a solvent in a solution is directly proportional to the mole fraction of the solvent in the solution. In this case, we can apply it to the ethane gas above the solution.

Understanding the Relationship

First, let's break down what we know:

• The partial pressure of ethane when the mass is 6.56 × 10–3 g is 1 bar.
• We want to find the partial pressure when the mass is increased to 5.00 × 10–2 g.

Calculating Moles of Ethane

To apply Raoult's Law, we need to convert the mass of ethane into moles. The molar mass of ethane (C₂H₆) is approximately 30.07 g/mol. We can calculate the number of moles for both scenarios:

For 6.56 × 10–3 g of ethane:

Number of moles = mass / molar mass = 6.56 × 10–3 g / 30.07 g/mol ≈ 0.000218 moles

For 5.00 × 10–2 g of ethane:

Number of moles = 5.00 × 10–2 g / 30.07 g/mol ≈ 0.00166 moles

Finding the Mole Fraction

Next, we need to find the mole fraction of ethane in the solution. The mole fraction (X) is calculated as:

X = moles of ethane / (moles of ethane + moles of solvent)

Assuming the solvent is water and using a typical mass of water (let's say 100 g for simplicity), we can calculate the moles of water:

Moles of water = 100 g / 18.015 g/mol ≈ 5.55 moles

Now, we can find the mole fraction for both cases:

For 6.56 × 10–3 g of ethane:

X₁ = 0.000218 / (0.000218 + 5.55) ≈ 0.000039

For 5.00 × 10–2 g of ethane:

X₂ = 0.00166 / (0.00166 + 5.55) ≈ 0.000299

Applying Raoult's Law

According to Raoult's Law, the partial pressure of a gas is proportional to its mole fraction:

P₁ = X₁ * P₀

P₂ = X₂ * P₀

Where P₀ is the vapor pressure of pure ethane. Since we know that P₁ = 1 bar when X₁ is calculated, we can express P₀ in terms of X₁:

P₀ = P₁ / X₁ = 1 bar / 0.000039 ≈ 25641 bar

Now, we can find P₂:

P₂ = X₂ * P₀ = 0.000299 * 25641 bar ≈ 7.67 bar

Final Result

The partial pressure of ethane over the solution containing 5.00 × 10–2 g of ethane is approximately 7.67 bar. This demonstrates how the increase in the amount of ethane in the solution significantly raises its partial pressure, illustrating the direct relationship between the amount of solute and its vapor pressure in a solution.