Pure ethylene glycol, , is added to 2.00 kg of water in the cooling system of a car. The vapor pressure of the water in the system when the temperature is 90 °C is 451 mm Hg. What mass of glycol was added? (Assume the solution is ideal and ethylene glycol is nonvolatile. The vapor pressure of water at 90 °C is 525.8 mm Hg.)

1367.7 g of ethylene glycol was added to the solution

Explanation:

In order to find out the mass of glycol we added, we apply the colligative property of lowering vapor pressure: ΔP = P° . Xm

ΔP = Vapor pressure of pure solvent (P°) - Vapor pressure of solution(P')

525.8 mmHg - 451 mmHg = 451 mmHg . Xm

74.8 mmHg / 451 mmHg = Xm → 0.166 (mole fraction of solute)

Xm = Mole fraction of solute / Moles of solute + Moles of solvent

We can determine the moles of solvent → 2000 g . 1  mol/18 g = 111.1 mol

(Notice we converted the 2kg of water to g)

0.166 = Moles of solute / Moles of solute + 111.1 moles of solvent

0.166 (Moles of solute + 111.1 moles of solvent) = Moles of solute

18.4 moles = Moles of solute - 0.166 moles of solute

18.4 = 0.834 moles of solute → Moles of solute = 18.4/0.834 = 22.06 moles

Let's convert the moles to mass → 62 g/mol . 22.06 mol = 1367.7 g

h let me think