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.

Answer 2

The vapour pressure of the solution can be calculated using Raoult's Law, which states that the vapour pressure of the solution is the sum of the partial pressures of the solvent and solute.

First, calculate the mole fraction of water (solvent):

• Moles of water = mass of water / molar mass of water =\(178.2 \, \text{g} / 18 \, \text{g/mol} = 9.9\  \text{mol}\)
• Moles of glucose = mass of glucose / molar mass of glucose = \(18 \, \text{g} / 180 \, \text{g/mol} = 0.1\  \text{mol}\)

Then, calculate the mole fraction of water and use Raoult's Law to find the vapour pressure of the solution:

\(P_{\text{solution}} = X_{\text{water}} \times P_{\text{water (pure)}}\)

The calculated vapour pressure is 752.4 torr.