AP Chemistry : Colligative Properties

Study concepts, example questions & explanations for AP Chemistry

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Example Question #11 : Colligative Properties

The vapor pressure of water is  at . Two moles of a nonvolatile solute are added to eight moles of water. What is the vapor pressure of this solution?

Possible Answers:

Correct answer:

Explanation:

A nonvolatile solute will not contribute to the vapor pressure of the solution, but will decrease the vapor pressure of the pure solvent. We can solve for the vapor pressure of the solution by using Raoult's law:

 

In other words, the new vapor pressure is equal to the molar fraction of the solvent multiplied by the vapor pressure of the pure solvent.

Example Question #12 : Colligative Properties

At room temperature, hexane has a vapor pressure of  and ethanol has a vapor pressure of . A solution of these two solvents at room temperature has a vapor pressure of . What percentage of the solution is ethanol?

Possible Answers:

Correct answer:

Explanation:

Since both of the solvents have a vapor pressure, we can find the percentage of ethanol in the solution by using Raoult's law, including both solvents in the equation:

Since we want to find the percentage of ethanol in the solution, we will designate the molar fraction of ethanol as . Since the sum of the molar fractions equals one, the molar fraction of hexane will be designated as .

So, 63.4% of the solution is ethanol.

Example Question #11 : Colligative Properties

A volatile solute with a vapor pressure of 23mmHg is added to a solvent with a vapor pressure of 85mmHg. What is the molar fraction of the solvent if the solution has a vapor pressure of 68mmHg?

Possible Answers:

Correct answer:

Explanation:

Since the solute is volatile, it will contribute to the total vapor pressure of the solution. As a result, we must incorporate it when solving for the total vapor pressure.

We can find the vapor pressures using Raoult's law.

Since we are solving for the molar fraction of the solvent, we will designate its molar fraction as . The sum of the molar fractions of each component must be equal to 1; thus, the molar fraction of the solute must be . Using these variables and the information given, we can solve for the molar fraction of the solvent.

Example Question #14 : Colligative Properties

What is the freezing point of a solution with  of sodium chloride in  of water?

 

Possible Answers:

Correct answer:

Explanation:

The equation for freezing point depression is , where  is the change in temperature,  is a constant related to the solvent,  is molality, and  is the van't Hoff factor, which is the number of ion particles from each dissolved molecule. We simply plug these numbers into the equation to find the new freezing point.

We know our constant is . Molality is moles of solute per kilogram of solution, and we know that the density of water is one kilogram per liter and the molecular weight of sodium chloride is .

The van't Hoff factor is . Sodium chloride creates only two ions when dissolved, one  and one .

Using these values, we can solve for the freezing point depression.

The freezing point will be decreased by . The normal freezing point is , making the new freezing point .

Example Question #15 : Colligative Properties

What is the melting point of a  aqueous solution that contains  of ?

Possible Answers:

Correct answer:

Explanation:

Since a salt has been added to the pure water, we can find the new melting point of the solution by using the freezing point depression equation:

The change in temperature is equal to the freezing point constant for the solvent multiplied by the molality of the solution multiplied by the van't Hoff factor. The van't Hoff factor is the number of ions that a salt will dissociate into when in solution.

For this particular salt, the van't Hoff factor is three. The molality will be equal to the moles of solute over the mass of the solvent.

Using these terms together in the original equation, we can find the freezing point depression.

This is the change in temperature from the regular freezing point. Since the freezing point for pure water is , the new melting point is .

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