Question:
18 g of glucose is dissolved in 178.2 g of water. The vapour pressure of the solution at 100$^\circ$C is (vapour pressure of pure water at 100$^\circ$C is 760 mm Hg)
18 g of glucose is dissolved in 178.2 g of water. The vapour pressure of the solution at 100$^\circ$C is (vapour pressure of pure water at 100$^\circ$C is 760 mm Hg)
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Non-volatile solutes always lower vapour pressure. Use mole fraction of solvent only.
Updated On: May 8, 2026
- $767.6$ mm Hg
- $760$ mm Hg
- $752.4$ mm Hg
- $725.4$ mm Hg
- $745.2$ mm Hg
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The Correct Option is C
Solution and Explanation
Concept:
Raoult’s law:
\[
P = X_{solvent} \cdot P^0
\]
Step 1: Moles calculation. Water: \[ \frac{178.2}{18} = 9.9 \text{ mol} \] Glucose: \[ \frac{18}{180} = 0.1 \text{ mol} \]
Step 2: Mole fraction of solvent. \[ X_{water} = \frac{9.9}{9.9 + 0.1} = 0.99 \]
Step 3: Vapour pressure. \[ P = 0.99 \times 760 = 752.4 \text{ mm Hg} \] \[ \boxed{752.4 \text{ mm Hg}} \]
Step 1: Moles calculation. Water: \[ \frac{178.2}{18} = 9.9 \text{ mol} \] Glucose: \[ \frac{18}{180} = 0.1 \text{ mol} \]
Step 2: Mole fraction of solvent. \[ X_{water} = \frac{9.9}{9.9 + 0.1} = 0.99 \]
Step 3: Vapour pressure. \[ P = 0.99 \times 760 = 752.4 \text{ mm Hg} \] \[ \boxed{752.4 \text{ mm Hg}} \]
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