What is the vapor pressure lowering D P of an aqueous glucose solution prepared by dissolving 9.01 grams of glucose, C₆H₁₂O₆, in 17.1 grams of water?
Molar Mass(C₆H₁₂O₆) = 180.16 g/mol and Molar Mass(H₂O) = 18.02 g/mol; vapor pressure of pure water = 23.8 torr (at 25 ° C)

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Quizplus · Accepted Answer

To calculate the vapor pressure lowering ($ \Delta P $), we use Raoult's law, which states that the vapor pressure lowering is directly proportional to the mole fraction of the solute in the solution. First, we calculate the moles of glucose and water:- Moles of glucose = mass / molar mass = $9.01 \, \text{g} / 180.16 \, \text{g/mol} = 0.05 \, \text{mol}$- Moles of water = mass / molar mass = $17.1 \, \text{g} / 18.02 \, \text{g/mol} = 0.95 \, \text{mol}$The mole fraction of glucose ($X_{\text{glucose}}$) is calculated as the moles of glucose divided by the total moles of the solution:$$X_{\text{glucose}} = \frac{0.05}{0.05 + 0.95} = \frac{0.05}{1} = 0.05$$The vapor pressure lowering ($ \Delta P $) is then the product of the mole fraction of the solute and the vapor pressure of the pure solvent:$$\Delta P = X_{\text{glucose}} \times \text{vapor pressure of pure water} = 0.05 \times 23.8 \, \text{torr} = 1.19 \, \text{torr}$$Rounding to two decimal places, $ \Delta P = 1.20 \, \text{torr}$.

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