A solution containing two volatile liquids A and B obeys Raoult's law. Derive the expression for its total vapour pressure.

📘 Derivation of Total Vapour Pressure

🔵 Step 1: Consider a Binary Solution

Consider a solution containing two volatile liquids A and B.

Let:

• x_A = Mole fraction of A in the liquid solution
• x_B = Mole fraction of B in the liquid solution
• P_A° = Vapour pressure of pure A
• P_B° = Vapour pressure of pure B
• P_A = Partial vapour pressure of A
• P_B = Partial vapour pressure of B

Since it is a binary solution:

x_A + x_B = 1

🟢 Step 2: Apply Raoult's Law to A

According to Raoult's law, the partial vapour pressure of a volatile component is equal to the product of its mole fraction and vapour pressure in the pure state.

P_A = x_AP_A°

🟠 Step 3: Apply Raoult's Law to B

P_B = x_BP_B°

🟣 Step 4: Calculate Total Vapour Pressure

The total vapour pressure of the solution is equal to the sum of the partial vapour pressures of A and B.

P_total = P_A + P_B

Substituting the Raoult's law expressions:

P_total = x_AP_A° + x_BP_B°

🎬 3D-Style Molecular Animation

Both volatile liquids contribute to the vapour pressure.

A B A B A B

A contributes → P_A = x_AP_A°

B contributes → P_B = x_BP_B°

Total → P_total = P_A + P_B

🔴 Step 5: Expression in Terms of Number of Moles

If n_A and n_B are the number of moles of A and B, then:

x_A = n_A / (n_A + n_B)

x_B = n_B / (n_A + n_B)

Therefore:

P_total = [n_AP_A° + n_BP_B°] / [n_A + n_B]

✅ Final Derived Expression

P_total = x_AP_A° + x_BP_B°

Thus, the total vapour pressure of a binary solution of two volatile liquids is the sum of their partial vapour pressures.

🎯 Exam Formula

P_A = x_AP_A°
P_B = x_BP_B°

P_total = x_AP_A° + x_BP_B°

A solution is prepared by dissolving 10 g of urea in 90 g of water. Calculate the mole fraction of urea.

📘 Complete Step-by-Step Solution

📌 Given Data

Mass of urea = 10 g

Mass of water = 90 g

Molar mass of urea = 60 g mol⁻¹

Molar mass of H₂O = 18 g mol⁻¹

🔵 Step 1: Calculate Moles of Urea

Moles = Mass ÷ Molar Mass

Moles of urea = 10 ÷ 60

= 0.1667 mol

🟢 Step 2: Calculate Moles of Water

Moles of water = 90 ÷ 18

= 5 mol

🟠 Step 3: Calculate Total Moles

Total moles = Moles of urea + Moles of water

= 0.1667 + 5

= 5.1667 mol

🟣 Step 4: Calculate Mole Fraction of Urea

Mole fraction of urea is the ratio of moles of urea to the total moles of the solution.

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χ_urea = Moles of urea ÷ Total moles

= 0.1667 ÷ 5.1667

≈ 0.03226

🎬 3D-Style Mole Fraction Animation

Urea molecules are present in a larger number of water molecules.

Urea H₂O H₂O H₂O H₂O H₂O

Urea = 0.1667 mol
Water = 5 mol

Total = 5.1667 mol

✅ Final Answer

Mole fraction of urea ≈ 0.0323

🎯 Important Formula

χ_solute = Moles of solute ÷ Total moles of solution

χ_urea + χ_water = 1