Question

18 g of glucose was dissolved in 360 g of water. The vapor pressure of the solution at 100°C will be

• A. 756.2 mm
• B. 755.2 mm
• C. 723.8 mm
• D. 76.0 mm

Hint:

Vapor pressure lowering depends on the mole fraction of the solute in the solution.

Answer 1

The Correct Option is B

Step 1: Moles of glucose \( = \dfrac{18}{180} = 0.1 \); moles of water \( = \dfrac{360}{18} = 20 \).
Step 2: Mole fraction of solute \( x_2 = \dfrac{0.1}{20 + 0.1} \), and by Raoult's law \( \dfrac{P_0 - P}{P_0} = x_2 \), with \( P_0 = 760 \) mm at 100 degrees C.
Step 3: Solve for \( P \) using this relative lowering.
\[ \boxed{P \approx 755.2 \ \text{mm}} \]

Answer 2

An alternative route is to work directly with the vapor pressure lowering \( \Delta P = P_0 \, x_{\text{solute}} \) and sanity-check the size of the answer against the dilution level, rather than solving for \( P \) first.

1. 756.2 mm: Comes very close, but from a slightly less precise handling of the mole fraction than the one that gives the final vapor pressure below.
2. 755.2 mm: Corresponds to a \( \Delta P \) of about 4.8 mm subtracted from 760 mm, the lowering that follows from the mole fraction of glucose in this dilute solution.
3. 723.8 mm: Implies a lowering of over 36 mm, roughly 8 to 10 times what a solute this dilute (about 0.5 mole percent) could plausibly cause.
4. 76.0 mm: Implies the vapor pressure collapsed to a tenth of its original value, which is not physically sensible for a mole fraction of solute under 1 percent.

Since the solute is present in only a small mole fraction, the vapor pressure lowering is a small correction to 760 mm.

Therefore, the correct answer is 755.2 mm.