Part A At 55.0 C, what is the vapor pressure of a solution prepared by dissolvin

Question

Part A

At 55.0 C, what is the vapor pressure of a solution prepared by dissolving 62.8 g of LiF in 265 g of water? The vapor pressure of water at 55.0 C is 118 mmHg. Assume complete dissociation of the solute.

Part B

The solvent for an organic reaction is prepared by mixing 60.0 mL of acetone (C3H6O) with 70.0 mLof 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?

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

Explanation / Answer

Part -A

The vapor pressure will be lowered by the non volatile solute, and the magnitude is related to the mole fraction of the solute. So, moles of solute = 62.8 g x 1 mol/26 g = 2.42 moles LiF
moles of particles or ions = 2.42 x 2 = 4.83 since LiF dissociates into 2 ions
moles water = 265 g x 1 mol/18g = 14.72 moles water
Total moles = 14.72 + 4.83 = 19.55 moles
mole fraction of water = 14.72/19.55 =0.7529
vapor pressure of solution = 0.7529 x 118 mm Hg = 88.84 mm Hg

Part -B

In order to solve this problem we will perform a series of steps that convert the given information into relevant pressures.

First, we will convert the two substances' volumes into grams using the provided densities.
Acetone: 60.0 mL x (0.791 g/mL) = 47.46 grams of Acetone
Ethyl Acetate: 70.0 mL x (0.900 g/mL) = 63 grams of Ethyl Acetate

Now we convert these quantities to moles by dividing the mass by the molar mass:
Acetone: 47.46 g / (58.08034 g/mol) = 0.8171 moles Acetone
Ethyl Acetate: 63 g / (88.1063 g/mol) = 0.7150 moles Ethyl Acetate

Now we determine the mole fraction for each substance:
Acetone: 0.8171 moles / (0.8171 moles + 0.7150 moles) = 0.5333
Ethyl Acetate: 0.7150 moles / (0.8171 moles + 0.7150 moles) = 0.4667

Finally, we apply Raoult's Law: multiply the mole fraction by the vapor pressure from the table:
Acetone: 0.5333 x 230.0 mmHg = 122.66 mmHg
Ethyle Acetate: 0.4667 x 95.38 mmHg = 44.51 mmHg

Add the two vapor pressures:
Answer: 122.66 mmHg + 44.51 mmHg = 167.17 mmHg

Answer with three sig figs: 167 mmHg

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