When a specific amount of acetone (C3H6O) is added to 100.0 g of pure water at 65°C, the vapor pressure of water over the solution is lowered by 1.576 kPa. Given the vapor pressure of water at 65°C is 25.022 kPa, what is the mass of acetone added?

mass of acetone added:

w acetone = 21.676 g

Explanation:

decrease in vapor pressure:

ΔP = - (P*a)(Xb,l)

∴ ΔP = - 1.576 KPa = Pa - P*a

∴ a: water (solvent)

∴ b: C3H6O (solute)

∴ P*a (65°C) = 25.022 KPa

⇒ ΔP/(- P*a) = Xb,l

⇒ Xb.l = - 1.576 KPa/(- 25.022 KPa) = 0.063

∴ Xb = nb/nt = nb/(na + nb) = (wb/Mb)/[(wa/Ma) + (wb/Mb)]

∴ Mb = 58.08 g/mol (molar mass acetone)

∴ Ma = 18.015 g/mol (molar mass water)

∴ wa = 100 g

⇒ 0.063 = (wb/58.08)/[(100/18.015) + (wb/58.08)]

⇒ (0.063)[(5.55 + (wb/58.08)] = wb/58.08

⇒ 0.3497 + 1.085 E-3wb = wb/58.08

⇒ 20.3106 + 0.063wb = wb

⇒ 20.3106 = wb - 0.063wb

⇒ 20.3106 = 0.937wb

⇒ 20.3106/0.937 = wb

⇒ 21.676 g = wb

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