AP Chemistry 3 Properties of Substances and Mixtures Questions

Explore AP Chemistry Unit 3 questions on intermolecular forces, solid and gas properties, solutions, separations, solubility, photons, and Beer-Lambert spectroscopy.

Covered syllabus points

3.1 Intermolecular and Interparticle Forces 3.2 Properties of Solids 3.3 Solids, Liquids, and Gases 3.4 Ideal Gas Law 3.5 Kinetic Molecular Theory 3.6 Deviation from Ideal Gas Law 3.7 Solutions and Mixtures 3.8 Representations of Solutions 3.9 Separation of Solutions and Mixtures 3.10 Solubility 3.11 Spectroscopy and the Electromagnetic Spectrum 3.12 Properties of Photons 3.13 Beer-Lambert Law

[Maximum number: 2]

A scientist is investigating the properties of a mixture of $\mathrm{CH}_{3} \mathrm{OH}$ and $\mathrm{H}_{2} \mathrm{CO}$ . The scientist generates the following diagram to represent the mixture.

(a)

In the diagram provided, draw a SINGLE dashed line (---) to represent a strong hydrogen-bonding attraction between one $\mathrm{CH}_{3} \mathrm{OH}$ molecule and one $\mathrm{H}_{2} \mathrm{CO}$ molecule in the mixture.

[ 1 ]

(b)

The scientist plans to cool a gaseous mixture of $\mathrm{CH}_{3} \mathrm{OH}$ and $\mathrm{H}_{2} \mathrm{CO}$ to form a liquid mixture and finds data on the two compounds. The data are summarized in the table.

[ 1 ]

(i)

Propose a temperature to which the mixture should be cooled such that $\mathrm{CH}_{3} \mathrm{OH}$ and $\mathrm{H}_{2} \mathrm{CO}$ will both be liquids.

[ 1 ]

For a correct proposal:

Point 03

The proposed temperature should be in the range $181 \mathrm{~K}-254 \mathrm{~K}$ .

[Maximum number: 3]

Complete Lewis diagrams and some physical properties for compounds X and Y are given.

(a)

A student claims that compound Y has a higher boiling point than that of compound X because compound Y has stronger London dispersion forces. Do you agree or disagree?

Justify your answer.

[ 1 ]

(b)

An equimolar mixture of the two compounds is heated. When the mixture reaches $82^{\circ} \mathrm{C}$ , which compound will have the higher vapor pressure? Justify your answer.

[ 1 ]

(c)

The mixture is heated to 198°C in a sealed, rigid 12.5 L container, at which point both substances are gases and the total pressure in the container is 2.30 atm. Calculate the number of moles of gas particles in the container.

[ 1 ]

	For the correct calculated value: \[ \begin{aligned}	P V=n R T
	n=\frac{P V}{R T}=\frac{(2.30 ~atm)(12.5 ~L)}{\left(0.08206 \frac{~L \cdot ~atm}{~K \cdot ~mol}\right)(471 ~K)}=0.744 ~mol \end{aligned} \]	Point 04

[Maximum number: 4]

A student investigates gas behavior using a rigid cylinder with a movable piston of negligible mass, as shown in the diagram above. The cylinder contains 0.325 mol of $\mathrm{O}_{2}(g)$ .

(a)

The cylinder has a volume of 7.95 L at 25°C and 1.00 atm. Calculate the density of the $\mathrm{O}_{2}(g)$ , in g/L, under these conditions.

[ 1 ]

For the correct calculated value: 1 point
Accept one of the following:

- $0.325 \mathrm{~mol} \mathrm{O}_{2} \times \frac{32.00 \mathrm{~g} \mathrm{O}_{2}}{1 \mathrm{~mol} \mathrm{O}_{2}}=10.4 \mathrm{~g} \mathrm{O}_{2}$

D=\frac{m}{V}=\frac{10.4 \mathrm{~g}}{7.95 \mathrm{~L}}=1.31 \mathrm{~g} / \mathrm{L}

- $D=\frac{m}{V}=\frac{P(M M)}{R T}=\frac{(1.0 \mathrm{~atm})(32.00 \mathrm{~g} / \mathrm{mol})}{\left(0.08206 \frac{\mathrm{~L} \cdot \mathrm{~atm}}{\mathrm{~mol} \cdot \mathrm{~K}}\right)(298 \mathrm{~K})}=1.31 \mathrm{~g} / \mathrm{L}$

(b)

Attempting to change the density of the $\mathrm{O}_{2}(g)$ , the student opens the valve on the side of the cylinder, pushes down on the piston to release some of the gas, and closes the valve again. The temperature of the gas remains constant at 25°C. Will this action change the density of the gas remaining in the cylinder? Justify your answer.

[ 1 ]

For the correct answer and a valid justification: 1 point
Accept one of the following:

- No, the density of the gas remains constant because P, R, and T remain constant AND the mass and volume of $O_{2}$ decrease proportionately.

- A mathematical justification is shown below.

D=\frac{m}{V}=\frac{n \text { moles of } \mathrm{O}_{2} \times \text { molar mass of } \mathrm{O}_{2}}{\frac{n R T}{P}}=\frac{P \times\left(\text { molar mass of } \mathrm{O}_{2}\right)}{R T}

(c)

The student tries to change the density of the $\mathrm{O}_{2}(g)$ by cooling the cylinder to -55°C, which causes the volume of the gas to decrease. Using principles of kinetic molecular theory, explain why the volume of the $\mathrm{O}_{2}(g)$ decreases when the temperature decreases to -55°C.

[ 1 ]

For a valid explanation: 1 point
Accept one of the following:

- As the gas cools, the average kinetic energy (speed) of the $O_{2}$ molecules decreases. The molecules rebound with less energy when they collide with each other and the walls of the container. The spacing between particles decreases, causing the volume occupied by the gas to decrease.

- As the gas cools, the average kinetic energy (speed) of the $O_{2}$ molecules decreases. The molecules rebound with less energy when they collide with each other and the walls of the container. The only way for the molecules to maintain a constant rate of collisions with the walls of the container (maintaining a pressure of 1.00 atm) is for the volume of the gas to decrease.

(d)

The student further cools the cylinder to $-180^{\circ} \mathrm{C}$ and observes that the measured volume of the $\mathrm{O}_{2}(g)$ is substantially smaller than the volume that is calculated using the ideal gas law. Assume all equipment is functioning properly. Explain why the measured volume of the $\mathrm{O}_{2}(g)$ is smaller than the calculated volume. (The boiling point of $\mathrm{O}_{2}(l)$ is $-183^{\circ} \mathrm{C}$ .)

[ 1 ]

[Maximum number: 1]

Answer the following questions about magnesium.

A student combines 35.00 mL of $1.85 \times 10^{-3} M \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}(a q)$ with 50.00 mL of $2.80 \times 10^{-4} M \mathrm{NaOH}(a q)$ , as shown in the diagram. Calculate $\left[\mathrm{Mg}^{2+}\right]$ after the two solutions are combined but before any reaction takes place. (Assume that volumes are additive.)

Beaker 3

For the correct calculated value:

Point 06

$\begin{aligned} & M_{1} V_{1}=M_{2} V_{2} \\ & M_{2}=\frac{\left(1.85 \times 10^{-3} M\right)(0.03500 \mathrm{~L})}{(0.03500 \mathrm{~L}+0.05000 \mathrm{~L})}=7.62 \times 10^{-4} M \end{aligned}$

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question (i)

Question 2

Question (a)

Question (b)

Question (c)

Question 3

Question (a)

Question (b)

Question (c)

Question (d)

Question 4