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: Recognise the property.
Glucose is non-volatile, so the solution's vapour pressure is lowered relative to pure water - a relative-lowering-of-vapour-pressure problem.
Step 2: State Raoult's law.
$\dfrac{P_1^0 - P_1}{P_1^0} = x_2$, the mole fraction of solute. So $P_1 = P_1^0(1 - x_2) = P_1^0 \, x_1$, where $x_1$ is the solvent mole fraction.
Step 3: Find moles.
Moles of glucose $= \dfrac{1.8}{180} = 0.01$. Moles of water $= \dfrac{16.2}{18} = 0.90$.
Step 4: Get the solvent mole fraction.
$x_1 = \dfrac{0.90}{0.90 + 0.01} = \dfrac{0.90}{0.91} = 0.989$.
Step 5: Multiply by pure vapour pressure.
$P_1 = 24 \times 0.989 = 23.74$ mm Hg.
Step 6: Round to the option.
$23.74 \approx 23.8$ mm Hg, which is option (4). The very small drop makes sense because the solution is extremely dilute.
\[ \boxed{P_{\text{solution}} \approx 23.8 \text{ mm Hg}} \]