When a non-volatile solute is added to the solvent, the vapour pressure of the solvent decreases by 10 mm of Hg. The mole fraction of the solute in the solution is 0.2. What would be the mole fraction of the solvent if the decrease in vapour pressure is 20 mm of Hg?
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Raoult’s law relates the decrease in vapour pressure to the mole fraction of the solute in the solution. The mole fraction of the solvent can be found by subtracting the solute’s mole fraction from 1.
To address this problem, we first establish Raoult's law: the vapor pressure of a solvent in a solution is directly proportional to the solvent's mole fraction. A non-volatile solute does not contribute to the vapor pressure.
Initially, a 10 mm Hg decrease in vapor pressure corresponds to a solute mole fraction of \(x_{\text{solute}} = 0.2\). Let \(P^0\) represent the initial vapor pressure of the pure solvent.
Raoult's law states: \(P_{\text{solution}} = x_{\text{solvent}} \times P^0\).
The given decrease in vapor pressure is 10 mm Hg, which means: \(P^0 - P_{\text{solution}} = 10 \text{ mm Hg}\).
From Raoult's law, the mole fraction of the solvent is: \(x_{\text{solvent}} = 1 - x_{\text{solute}} = 1 - 0.2 = 0.8\).
Now, consider a change in vapor pressure to 20 mm Hg. Applying Raoult's law again:
\(P^0 - P_{\text{solution}} = 20 \text{ mm Hg}\).
Let \(x'_{\text{solvent}}\) denote the new mole fraction of the solvent.
Using the proportional decrease relationship:
\(x'_{\text{solvent}} \times P^0 = P^0 - 20\).
Since the total pressure decrease is doubled, the new mole fraction of the solute, \(x_{\text{solute}}'\), will be inversely related to the solvent coverage.
