Question

Derive relationship between relative lowering of vapour pressure and molar mass of non volatile solute.

Hint:

This formula is valid only for dilute solutions of non-electrolytes. If the solute dissociates or associates, the van't Hoff factor (\(i\)) must be included.

Answer 1

Step 1: Understanding the Concept:
According to Raoult's law, the relative lowering of vapour pressure for a solution containing a non-volatile solute is equal to the mole fraction of the solute in the solution.
Step 2: Key Formula or Approach:
The relative lowering of vapour pressure is given by:
\[ \frac{P_1^0 - P_1}{P_1^0} = x_2 \]
Where \(P_1^0\) is the vapour pressure of the pure solvent, \(P_1\) is the vapour pressure of the solution, and \(x_2\) is the mole fraction of the solute.
Step 3: Detailed Explanation:
Let \(n_1\) and \(n_2\) be the number of moles of solvent and solute respectively.
The mole fraction of solute is:
\[ x_2 = \frac{n_2}{n_1 + n_2} \]
For dilute solutions, the number of moles of solute \(n_2\) is very small compared to the number of moles of solvent \(n_1\).
Thus, \(n_1 + n_2 \approx n_1\).
Substituting this into the Raoult's law expression:
\[ \frac{P_1^0 - P_1}{P_1^0} \approx \frac{n_2}{n_1} \]
We know that \(n_2 = \frac{W_2}{M_2}\) and \(n_1 = \frac{W_1}{M_1}\), where \(W\) is mass and \(M\) is molar mass.
Substituting these values:
\[ \frac{P_1^0 - P_1}{P_1^0} = \frac{W_2 / M_2}{W_1 / M_1} = \frac{W_2 \cdot M_1}{M_2 \cdot W_1} \]
Step 4: Final Answer:
The final relationship is \( M_2 = \frac{W_2 \cdot M_1 \cdot P_1^0}{W_1 \cdot (P_1^0 - P_1)} \).