20 g of a non-volatile solute is dissolved in 180 g of water. The vapour pressure of the solution is 23.5 mm Hg while that of pure water is 24 mm Hg. Calculate the molar mass of the solute.

📘 Complete Step-by-Step Solution

📌 Given Data

Mass of solute = 20 g

Mass of water = 180 g

Vapour pressure of pure water, P° = 24 mm Hg

Vapour pressure of solution, P = 23.5 mm Hg

Molar mass of water = 18 g mol⁻¹

🔵 Step 1: Apply Raoult's Law

For a solution containing a non-volatile solute, the relative lowering of vapour pressure is equal to the mole fraction of the solute.

(P° − P) / P° = X_solute

X_solute = (24 − 23.5) / 24

= 0.5 / 24

= 0.020833

🟢 Step 2: Calculate Moles of Water

Moles of water = Mass / Molar mass

= 180 / 18

= 10 mol

🟠 Step 3: Find Moles of Solute

We know:

X_solute = n_solute / (n_solute + n_water)

Let the moles of solute be n.

0.020833 = n / (n + 10)

Cross-multiplying:

0.020833(n + 10) = n

0.020833n + 0.20833 = n

0.20833 = 0.979167n

n ≈ 0.2128 mol

🟣 Step 4: Calculate Molar Mass of Solute

Molar mass = Mass / Number of moles

M = 20 / 0.2128

M ≈ 94.0 g mol⁻¹

🎬 3D-Style Vapour Pressure Animation

The non-volatile solute reduces the escaping tendency of water molecules.

H₂O H₂O Solute H₂O H₂O

Pure water: 24 mm Hg

Solution: 23.5 mm Hg

Lowering = 0.5 mm Hg

✅ Final Answer

Molar Mass of Solute ≈ 94 g mol⁻¹

🎯 Important Formula

(P° − P) / P° = X_solute

X_solute = n_solute / (n_solute + n_solvent)

Molar Mass = Mass / Moles