Answer:
Vapor pressure of solution = 23.9 Torr
Explanation:
Let's apply the colligative poperty of vapor pressure to solve this:
ΔP = P° . Xm
ΔP = Vapor pressure of pure solvent - Vapor pressure of solution
We have solvent and solute mass, so let's find out the moles of each.
55.3 g / 62 g/mol = 0.89 moles
285.2 g / 18 g/mol = 15.84 moles
Let's determine the mole fraction of ethylene glycol.
Mole fraction = Moles of ethylene glyco / Total moles
0.89 moles / (0.89 + 15.84) = 0.053
25.3 Torr - Vapor pressure of solution = 25.3 Torr . 0.053
Vapor pressure of solution = 25.3 Torr . 0.053 - 25.3 Torr
Vapor pressure of solution = 23.9 Torr
Answer:
P (25°C) = 22.523 torr
Explanation:
low volatile solute and diluted solution:
∴ a: water
∴ b: C2H6O2
∴ P*a (25°C) = 23.8 torr
∴ wb = 55.3 g
∴ Mb = 62.07 g/mol
∴ wa = 285.2 g
∴ Ma = 18.015 g/mol
⇒ Xb = (55.3/62.07) / ((285.2/18.015) + (55.3/62.07))
⇒ Xb = 0.0536
⇒ P (25°C) = P*a - Xb.P*a
⇒ P (25°C) = 23.8 torr - ((0.0536)(23.8 torr))
⇒ P (25°C) = 22.523 torr
