What is the vapor pressure of the solution if 15.0 g of water is dissolved in 100.0 g of ethyl alcohol at 25 ∘C? The vapor pressure of pure water is 23.8 mmHg, and the vapor pressure of ethyl alcohol is 61.2 mmHg at 25 ∘C.

44.2 mmHg

Explanation:

We have to apply the colligative property of lowering vapor pressure:

P° - P' = P°. Xm

where P' refers to vapor pressure of solution

P° refers to vapor pressure of pure solvent

Xm is the mole fraction of solute

Let's determine the mole fraction (moles of solute / Total moles)

Total moles = Moles of solute + Moles of solvent

Moles of solute → 15 g. 1 mol/18g = 0.833 moles

Moles of solvent → 100 g. 1mol / 46 g = 2.174 moles

Total moles = 0.833 + 2.174 = 3.007

Xm = 0.833 / 3.007 = 0.277

We replace data in the formula: 61.2 mmHg - P' = 61.2 mmHg. 0.277

P' = - (61.2 mmHg. 0.277 - 61.2 mmHg) → 44.2 mmHg