Question

The vapour pressure of H\(_2\)O at 323K is 95 mm of Hg. 176g of sucrose (Molar mass = 342 gmol\(^{-1}\)) is added to 900g of H\(_2\)O at 323K. 
The vapour pressure of solution is about

• A. 93.94 mm
• B. 92.88 mm
• C. 96.06 mm
• D. 95.33 mm
• E. 94.06 mm

Hint:

Raoult's law helps calculate the vapour pressure of a solution based on the mole fractions of the solvent and solute.

Answer 1

Answer: (E)

Using Raoult's law: \[ P_{{solution}} = P_{{solvent}} \times \frac{n_{{solvent}}}{n_{{solvent}} + n_{{solute}}} \] First, calculate the moles of H\(_2\)O and sucrose: \[ {moles of H}_2{O} = \frac{900}{18} = 50 \, {mol} \] \[ {moles of sucrose} = \frac{176}{342} = 0.514 \, {mol} \] Now calculate the vapour pressure of the solution: \[ P_{{solution}} = 95 \, {mm Hg} \times \frac{50}{50 + 0.514} \approx 94.06 \, {mm Hg} \] Thus, the correct answer is (E).