Determine the vapour pressure of a solution at 25 °C that contains 76.6 g of glucose (C₆H₁₂O₆) in 250.0 mL of water. The vapour pressure of pure water at 25 °C is 0.03173 bar. A) 0.360 bar B) 0.318 bar C) 0.0308 bar D) 0.0387 bar E) 0.00567 bar

Question

Quizplus · Accepted Answer

The vapor pressure of the solution is calculated using Raoult's Law. First, find the mole fraction of water in the solution. The moles of glucose are 76.6 g / 180.18 g/mol = 0.425 mol. The moles of water are 250 g / 18.015 g/mol = 13.88 mol. The mole fraction of water is 13.88 / (13.88 + 0.425) = 0.970. The vapor pressure of the solution is 0.970 * 0.03173 bar = 0.0308 bar.

Determine the Vapour Pressure of a Solution at 25 °C