Question

What is vapour pressure of solution containing 1.8 g glucose in 16.2 g water? ($P_1^0 = 24$ mm Hg and Molar mass of glucose = 180 g mol$^{-1}$)

• A. 18.1 mm Hg
• B. 15.7 mm Hg
• C. 12.4 mm Hg
• D. 23.8 mm Hg

Hint:

Since a small amount of glucose (1.8 g) is added to water, the vapour pressure lowering will be very small. You can immediately eliminate options (A), (B), and (C) because their values are significantly below the pure solvent's vapour pressure of 24 mm Hg.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:

The question asks for the vapour pressure of an aqueous solution containing a non-volatile solute (glucose) dissolved in water. We are given the mass of glucose, mass of water, molar mass of glucose, and the vapour pressure of pure water.


Step 2: Key Formula or Approach:

According to 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: $$\frac{P_1^0 - P_1}{P_1^0} = X_2 = \frac{n_2}{n_1 + n_2}$$ where $P_1^0$ is the vapour pressure of pure solvent, $P_1$ is the vapour pressure of the solution, $n_2$ is the moles of solute (glucose), and $n_1$ is the moles of solvent (water). For dilute solutions, it can be approximated as: $$\frac{P_1^0 - P_1}{P_1^0} \approx \frac{W_2 \times M_1}{M_2 \times W_1}$$ where $W$ represents mass and $M$ represents molar mass.


Step 3: Detailed Explanation:

Given values: Mass of glucose ($W_2$) = 1.8 g, Molar mass of glucose ($M_2$) = 180 g mol$^{-1}$ Mass of water ($W_1$) = 16.2 g, Molar mass of water ($M_1$) = 18 g mol$^{-1}$ Vapour pressure of pure water ($P_1^0$) = 24 mm Hg Let's find the number of moles: $$n_2 = \frac{1.8}{180} = 0.01\text{ mol}$$ $$n_1 = \frac{16.2}{18} = 0.90\text{ mol}$$ Now substitute the total mole fractions accurately: $$\frac{24 - P_1}{24} = \frac{0.01}{0.90 + 0.01} = \frac{0.01}{0.91} \approx 0.011$$ $$24 - P_1 = 24 \times 0.011 = 0.264\text{ mm Hg}$$ $$P_1 = 24 - 0.264 = 23.736\text{ mm Hg} \approx 23.8\text{ mm Hg}$$

Step 4: Final Answer:
The vapour pressure of the solution is approximately 23.8 mm Hg, which corresponds to option (D).