Question

Vapour pressure of a solution containing 18 g of glucose and 178.2 g of water at 100°C is :
(Vapour pressure of pure water at 100°C = 760 torr)

• A. 76.0 torr
• B. 752.4 torr
• C. 7.6 torr
• D. 3207.6 torr

Answer 1

The Correct Option is B

We can use Raoult's Law to calculate the vapor pressure of the solution:

P_solution = P_{pure solvent} × X_solvent

Where:

• P_solution is the vapor pressure of the solution.
• P_{pure solvent} is the vapor pressure of the pure solvent (water in this case).
• X_solvent is the mole fraction of the solvent (water) in the solution.

Step 1: Calculate the moles of glucose (C₆H₁₂O₆) and water (H₂O)

• Moles of glucose (C₆H₁₂O₆): Moles of glucose = \(\frac {mass\ of\ glucose}{molar\ mass\ of\ glucose}\) = \(\frac {18}{180}= 0.1\ mol\)
• Moles of water (H₂O): Moles of water =\(\frac {mass\ of\ water}{molar\ mass\ of\ water}\) = \(\frac {178.2}{18}\) g/mol = \(9.9\) mol

Step 2: Calculate the mole fraction of water (X_water)

X_water = moles of water / (moles of water + moles of glucose) = \(\frac {9.9}{(9.9 + 0.1)}= \frac {9.9}{10} = 0.99\)

Step 3: Apply Raoult's Law

P_solution = P_{pure solvent} × X_solvent P_solution = 760 torr × 0.99 P_solution = 752.4 torr

The vapor pressure of the solution is 752.4 torr, so the correct answer is (B) 752.4 torr.