i just need help with part B please background info As solut

*i just need help with part B please

(background info)

As solute is dissolved in a solvent, the vapor pressure of the solution changes according to Raoult\'s law

Psoln=Psolv×Xsolv

where Psoln is the vapor pressure of the solution, Psolv is the vapor pressure of the pure solvent, and Xsolv is the mole fraction of the solvent. If the solute dissociates into ions, the term Xsolv must be modified to take into consideration the total number of moles of particles in the solution, both ions and molecules. When a solution contains two volatile components, A and B, the total pressure of the solution is equal to the sum of the individual vapor pressures according to Dalton\'s law as follows: Ptotal=PA×XA +PB×XB

(end background info)

At 55.0 C, what is the vapor pressure of a solution prepared by dissolving 60.2 g of LiF in 259 g of water? The vapor pressure of water at 55.0 C is 118 mmHg. Assume complete dissociation of the solute.

Express your answer to three significant figures and include the appropriate units.

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Correct

Significant Figures Feedback: Your answer 88.95mmHg was either rounded differently or used a different number of significant figures than required for this part.

Part B

The solvent for an organic reaction is prepared by mixing 70.0 mL of acetone

(C3H6O) with 45.0 mL of 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

Express your answer to three significant figures and include the appropriate units.

Solution

Pure acetone @ 25.0 degC:
70.0g, 230.0 mmHg, 0.791 g/mL

C3H6O
3 C * 12.0107 = 36.032
6 H * 1.00794 = 6.048
1 O * 15.9994 = 15.999
MW of acetone = 58.079g acetone/1mol acetone

mol acetone = (70.0g acetone) x (1mol acetone/58.079g acetone)

(70)*(0.01721792730591091444411921692867)

= 1.2052549 mol acetone

Pure ethyl acetate @ 25.0 degC:
45.0g, 95.38 mmHg, 0.900 g/mL

C4H8O2
4 C * 12.0107 = 48.043
8 H * 1.00794 = 8.064
2 O * 15.9994 = 31.999
MW ethyl acetate = 88.105g ethyl acetate/1mol ethyl acetate

mol ethyl acetate = (45.0g ethyl acetate)
x (1mol ethyl acetate/88.105g ethyl acetate)

(45) x (0.0113500936382725157)
= 0.51075421372 mol ethyl acetate

total moles solvent
= 1.2052549 mol acetone + 0.51075421372 mol ethyl acetate

= 1.716009113722263 mol solvent

Equation for partial pressures:
Ppartial = Ppure x (mol pure/total moles)

Ppartial acetone = Ppure acetone x (mol acetone/mol solvent)

= (230.0 mmHg) x (1.2052549 mol acetone/1.716009113 mol solvent)
=(230) x (0.7023592653846238635913)
Ppartial acetone = 161.54263103846 mmHg

Ppartial ethyl acetate = Ppure ethyl acetate x (mol ethyl acetate/mol solvent)

= (95.38 mmHg) x (0.51075421372 mol ethyl acetate/1.716009113 mol solvent)

=(95.38) x (0.29764143857)

Ppartial ethyl acetate = 28.388973100867 mmHg

Ptotal = sum of partial pressures
= Ppartial acetone + Ppartial ethyl acetate

= 161.5426310 mmHg + 28.388973100867 mmHg

Ptotal = 189.9315731008672096 mmHg

Answer: 190 mm Hg
(at 3 sig.figs.)