Question

What is the approximate vapor pressure of an isopropanol solution containing 9.2 g of a non volatile solute, glycerol ($C_{3}H_{8}O_{3}$) and 60 g of isopropanol ($C_{3}H_{8}O$) at $82^{\circ}C$? Given vapor pressure of pure isopropanol at $82^{\circ}C$ is 100 kPa.

• A. 81 kPa
• B. 91 kPa
• C. 101 kPa
• D. 111 kPa

Hint:

Vapor pressure always drops when you add a non-volatile solute (Relative Lowering of Vapor Pressure).

Answer 1

The Correct Option is B

Step 1: Concept
According to Raoult's Law, the vapor pressure of a solution ($P_s$) is equal to the mole fraction of the solvent ($\chi_{solv}$) multiplied by the vapor pressure of the pure solvent ($P^{\circ}$).

Step 2: Meaning
Moles of glycerol = $9.2 / 92 = 0.1$ mol. Moles of isopropanol = $60 / 60 = 1.0$ mol.

Step 3: Analysis
Mole fraction of isopropanol = $1.0 / (1.0 + 0.1) = 1.0 / 1.1 \approx 0.909$. Vapor pressure of solution = $0.909 \times 100 \text{ kPa} = 90.9 \text{ kPa}$.

Step 4: Conclusion
The vapor pressure of the solution is approximately 91 kPa. Final Answer: (B)