At 55.0 ?C, what is the vapor pressure of a solution prepared by dissolving 70.6

Question

At 55.0 ?C, what is the vapor pressure of a solution prepared by dissolving 70.6g of LiF in 265g of water? The vapor pressure of water at 55.0 ?C is 118 mmHg. Assume complete dissociation of the solute.

The solvent for an organic reaction is prepared by mixing 60.0mL of acetone (C3H6O) with 59.0mLof ethyl acetate (C4H8O2). This mixture is stored at 25.0 ?C. The vapor pressure and the densities for the two pure components at 25.0 ?C are given in the following table. What is the vapor pressure of the stored mixture?

Compound Vapor pressure
(mmHg) Density
(g/mL) acetone 230.0 0.791 ethyl acetate 95.38 0.900

Explanation / Answer

1st part

According to Raoult's law

P = xP0

where P is the vapor pressure of solution and P0 for pure solvent. x is the mole fraction of the solvent

Now molar mass of LiF is 25.94, therefore no. of moles of LiF = 70.6/ 25.94 = 2.723 mol

No. of moles of water = 285/ 18 = 14.722 mol

so total no. of moles = 14.722 + 2.723 = 17.445 mol

therefore mole fraction of water, x = 14.722/ 17.455 = 0.844

therefore

P = xP0 = = 0.844 x 118 = 99.59 mm of Hg

therefore loweing of vapor pressure = 118 - 99.59 = 18.41 mm of Hg

complete dissociation is taking place, therefore vant Hof factor = 2

therefore acual lowering of vapor pressure = 2 x 18.41 = 36.82 mm of Hg

therefore vapor pressure of the solution is 118 - 36.82 = 81.18 mm of Hg

2nd part

mass of acetone = 60 x 0.791 g

molar mass of acetone = 58.08

Therefore no. of moles of acetone in the mixture = 60 x 0.791 / 58.08 = 0.817 mol

mass of ethyl acetate = 59 x 0.900 g

molar mass of ethyl acetate = 88.1

Therefore no. of moles of ethyl acetate in the mixture = 59 x 0.900 / 88.1 = 0.603 mol

Therefore total no. of moles = 0.817 + 0.603 = 1.42 mol

Therefore mole fraction of acetone = 0.817/ 1.42 = 0.575

and mole fraction of ethyl acetate = 0.603/ 1.42 = 0.425

therefore partial vapor pressure of acetone in the mixture

= mole fraction of acetone x vapor pressure of pure acetone = 0.575 x 230.0 = 132.25 mm of Hg

And partial vapor pressure of ethyl acetate in the mixture

= mole fraction of ethyl acetate x vapor pressure of pure ethyl acetate = 0.425 x 95.38 = 40.54 mm of Hg

Hence the vapor pressure of the stored mixture at 25oC

= partial vapor pressure of acetone + partial vapor pressure of ethyl acetate = 132.25 + 40.54 = 172.79 mm of Hg

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.