The Henry's law constant for pentan-3-one, C₂H₅COC₂H₅, in aqueous solution may be expressed in terms of concentration as 19.6 mol dm^-³ kPa^-1. Calculate the vapour pressure of pentan-3-one above an aqueous solution that contains 5.23 g of pentan-3-one in 250 cm³ of solution. A) 12.4 kPa B) 51.0 kPa C) 31.1 kPa D) 3.00 kPa

Question

Quizplus · Accepted Answer

First, calculate the molarity of pentan-3-one: Molar mass = 86.13 g/mol, so moles = 5.23 g / 86.13 g/mol = 0.0607 mol. Molarity = 0.0607 mol / 0.250 L = 0.2428 mol/L. Using Henry's law: P = C / k = 0.2428 mol/L / 19.6 mol/L kPa = 12.4 kPa.

The Henry's Law Constant for Pentan-3-One, C2H5COC2H5, in Aqueous Solution

The Henry's Law Constant for Pentan-3-One, C₂H₅COC₂H₅, in Aqueous Solution