CHEM 121 Test 4 Review – Step-by-Step Study Guidance

Q1. Barometric pressures are reported in inches of mercury (in. Hg). On a beautiful summer day in Chicago, the barometric pressure is 30.45 in. Hg. Convert this pressure to torr.

Background

Topic: Pressure Unit Conversions

This question tests your ability to convert between different units of pressure, specifically from inches of mercury (in. Hg) to torr.

Key Terms and Formulas:

• Pressure: The force exerted per unit area.

• 1 atm = 760 torr = 29.92 in. Hg

Step-by-Step Guidance

1. Write down the given pressure: 30.45 in. Hg.

2. Set up the conversion factor between in. Hg and torr. Recall that $1\ \text{atm} = 760\ \text{torr} = 29.92\ \text{in. Hg}$.

3. Write the conversion as a multiplication: $30.45\ \text{in. Hg} \times \frac{760\ \text{torr}}{29.92\ \text{in. Hg}}$.

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Q2. A fixed quantity of gas at a constant temperature exhibits a pressure of 737 torr and occupies a volume of 20.5 L. Calculate the volume the gas will occupy if the pressure is increased to 1.80 atm.

Background

Topic: Boyle's Law (Gas Laws)

This question tests your understanding of Boyle's Law, which relates the pressure and volume of a gas at constant temperature.

Key Terms and Formulas:

• Boyle's Law: $P_1V_1 = P_2V_2$

• Pressure must be in the same units for both $P_1$ and $P_2$.

Step-by-Step Guidance

1. List the known values: $P_1 = 737\ \text{torr}$, $V_1 = 20.5\ \text{L}$, $P_2 = 1.80\ \text{atm}$.

2. Convert $P_1$ to atm so both pressures are in the same units: $P_1 = 737\ \text{torr} \times \frac{1\ \text{atm}}{760\ \text{torr}}$.

3. Set up Boyle's Law: $P_1V_1 = P_2V_2$.

4. Rearrange to solve for $V_2$: $V_2 = \frac{P_1V_1}{P_2}$.

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Q3. A gas at a constant pressure has a temperature of 22°C and a volume of 55.60 L. Determine the volume of the gas if the temperature is increased to 38°C.

Background

Topic: Charles's Law (Gas Laws)

This question tests your ability to apply Charles's Law, which relates the volume and temperature of a gas at constant pressure.

Key Terms and Formulas:

• Charles's Law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$

• Temperature must be in Kelvin.

Step-by-Step Guidance

1. List the known values: $V_1 = 55.60\ \text{L}$, $T_1 = 22^\circ\text{C}$, $T_2 = 38^\circ\text{C}$.

2. Convert temperatures to Kelvin: $T(K) = T(^\circ\text{C}) + 273.15$.

3. Set up Charles's Law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.

4. Rearrange to solve for $V_2$: $V_2 = V_1 \times \frac{T_2}{T_1}$.

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Q4. A balloon originally had a volume of 4.39 L at 44°C and a pressure of 729 torr. To what temperature must the balloon be cooled to reduce its volume to 3.78 L if the pressure is constant?

Background

Topic: Charles's Law (Gas Laws)

This question tests your ability to use Charles's Law to relate volume and temperature at constant pressure.

Key Terms and Formulas:

• Charles's Law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$

• Temperature must be in Kelvin.

Step-by-Step Guidance

1. List the known values: $V_1 = 4.39\ \text{L}$, $T_1 = 44^\circ\text{C}$, $V_2 = 3.78\ \text{L}$.

2. Convert $T_1$ to Kelvin: $T_1 = 44 + 273.15$.

3. Set up Charles's Law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.

4. Rearrange to solve for $T_2$: $T_2 = T_1 \times \frac{V_2}{V_1}$.

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Q5. Calculate the volume of O2(g) needed at 500°C and 10.0 atm to react with 1.00 kg of CH4(g). 2CH4(g) + 1/2O2(g) → C2H6(g) + H2O(g)

Background

Topic: Gas Law Stoichiometry (Ideal Gas Law and Reaction Stoichiometry)

This question tests your ability to combine stoichiometry with the ideal gas law to determine the volume of a gas needed for a reaction.

Key Terms and Formulas:

• Ideal Gas Law: $PV = nRT$

• Mole-to-mole ratios from balanced chemical equations

• Molar mass of CH4: 16.0 g/mol

• R (gas constant): $0.0821\ \text{L·atm·mol}^{-1}\text{K}^{-1}$

Step-by-Step Guidance

1. Convert 1.00 kg CH4 to grams, then to moles using its molar mass.

2. Use the stoichiometric coefficients from the balanced equation to find moles of O2 needed.

3. Use the ideal gas law to solve for volume: $V = \frac{nRT}{P}$, where $n$ is the moles of O2, $T$ is in Kelvin, and $P$ is in atm.

4. Convert 500°C to Kelvin: $T = 500 + 273.15$.

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Q6. If CO2 and O2 are placed in a vessel with a pinhole, which gas do you expect to effuse first and how much faster does it effuse?

Background

Topic: Graham's Law of Effusion

This question tests your understanding of how the rate of effusion of a gas depends on its molar mass.

Key Terms and Formulas:

• Graham's Law: $\frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}}$

• $M_1$ and $M_2$ are the molar masses of the two gases.

Step-by-Step Guidance

1. Identify the molar masses: $M_{CO_2} = 44.0\ \text{g/mol}$, $M_{O_2} = 32.0\ \text{g/mol}$.

2. Set up Graham's Law to compare the rates of effusion: $\frac{\text{Rate}_{O_2}}{\text{Rate}_{CO_2}} = \sqrt{\frac{44.0}{32.0}}$.

3. Determine which gas effuses faster (the one with the lower molar mass).

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Q7. Identify the types of intermolecular forces that are present in C3H8 and CH3OCH3 and select the substance that has the highest boiling point.

Background

Topic: Intermolecular Forces and Boiling Points

This question tests your ability to identify intermolecular forces and relate them to boiling points.

Key Terms and Concepts:

• London Dispersion Forces (LDF): Present in all molecules, especially nonpolar ones.

• Dipole-Dipole Forces: Present in polar molecules.

• Hydrogen Bonding: Present when H is bonded to N, O, or F.

Step-by-Step Guidance

1. Determine the molecular structure and polarity of C3H8 (propane) and CH3OCH3 (dimethyl ether).

2. Identify which intermolecular forces are present in each compound.

3. Recall that stronger intermolecular forces generally lead to higher boiling points.

4. Compare the two substances and predict which has the higher boiling point based on the types of forces present.

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Q8. In which case can hydrogen bonding not occur between like molecules? NH3, (CH3)2NH, CH4, CH3COOH

Background

Topic: Hydrogen Bonding

This question tests your understanding of the requirements for hydrogen bonding between molecules.

Key Terms and Concepts:

• Hydrogen Bonding: Occurs when H is bonded to N, O, or F.

• Like molecules: Identical molecules interacting with each other.

Step-by-Step Guidance

1. Examine each molecule to see if it contains H bonded to N, O, or F.

2. Determine if the molecular structure allows for hydrogen bonding between like molecules.

3. Identify the molecule that cannot form hydrogen bonds with itself.

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Q9. The reason that some insects can walk on water is due to what property of liquids? What type of intermolecular force do you need for this property?

Background

Topic: Surface Tension and Intermolecular Forces

This question tests your understanding of surface tension and the intermolecular forces responsible for it.

Key Terms and Concepts:

• Surface Tension: The energy required to increase the surface area of a liquid.

• Hydrogen Bonding: Strong intermolecular force responsible for high surface tension in water.

Step-by-Step Guidance

1. Recall which property of liquids allows insects to walk on water (surface tension).

2. Identify the intermolecular force responsible for high surface tension in water.

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Q10. The enthalpy of fusion of water is 6.0 kJ/mol and the heat capacity is 75 J/mol·°C. How many kJ of heat would it take to convert 50. g of ice at 0°C to liquid water at 22°C? (Hvap = 38.0 kJ/mol, Cs = 4.184 J/g°C)

Background

Topic: Phase Changes and Heating Curves

This question tests your ability to calculate the total heat required for a phase change and temperature increase.

Key Terms and Formulas:

• Enthalpy of Fusion ($\Delta H_{fus}$): Heat required to melt 1 mol of a substance.

• Heat Capacity ($C_s$): Amount of heat needed to raise the temperature of 1 g of a substance by 1°C.

• Heat for melting: $q_1 = n \times \Delta H_{fus}$

• Heat for warming: $q_2 = m \times C_s \times \Delta T$

Step-by-Step Guidance

1. Calculate the moles of ice: $n = \frac{50.0\ \text{g}}{18.0\ \text{g/mol}}$.

2. Calculate the heat required to melt the ice at 0°C: $q_1 = n \times 6.0\ \text{kJ/mol}$.

3. Calculate the heat required to warm the resulting water from 0°C to 22°C: $q_2 = 50.0\ \text{g} \times 4.184\ \text{J/g}^\circ\text{C} \times (22 - 0)^\circ\text{C}$.

4. Convert $q_2$ from J to kJ if necessary.

5. Add $q_1$ and $q_2$ to get the total heat required.

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Q11. What is the normal boiling point for the substance shown in the phase diagram?

Background

Topic: Phase Diagrams

This question tests your ability to interpret a phase diagram and identify the normal boiling point.

Key Terms and Concepts:

• Normal Boiling Point: The temperature at which the vapor pressure equals 1 atm.

• Phase Diagram: Graph showing the state of a substance at various temperatures and pressures.

Step-by-Step Guidance

1. Locate the line separating the liquid and gas phases on the phase diagram.

2. Find the temperature where this line intersects 1 atm pressure.

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Q12. Which of the following in each pair is more likely to be soluble in hexane, C6H14? (i) CH3CH2CH2CH3 or CH3CH2OH (ii) CCl4 or CaCl2 (iii) C6H6 or C6H5OH (phenol)

Background

Topic: Solubility and Intermolecular Forces

This question tests your understanding of the "like dissolves like" principle and the role of polarity in solubility.

Key Terms and Concepts:

• Hexane: Nonpolar solvent.

• "Like dissolves like": Nonpolar solutes dissolve in nonpolar solvents; polar solutes dissolve in polar solvents.

Step-by-Step Guidance

1. Identify the polarity of each compound in the pairs.

2. Recall that nonpolar compounds are more soluble in nonpolar solvents like hexane.

3. Select the more nonpolar compound in each pair as the one more likely to be soluble in hexane.

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Q13. The Henry's law constant for CO2 is 3.1 x 10-2 mol/L-atm at 25°C. What pressure would be necessary in order to have a 0.25 M solution?

Background

Topic: Henry's Law (Gases in Solution)

This question tests your ability to use Henry's Law to relate the solubility of a gas to its pressure above a solution.

Key Terms and Formulas:

• Henry's Law: $C = k_H P$

• $C$ = concentration of dissolved gas (mol/L)

• $k_H$ = Henry's law constant (mol/L·atm)

• $P$ = partial pressure of the gas (atm)

Step-by-Step Guidance

1. Write the Henry's Law equation: $C = k_H P$.

2. Rearrange to solve for $P$: $P = \frac{C}{k_H}$.

3. Plug in the values: $C = 0.25\ \text{mol/L}$, $k_H = 3.1 \times 10^{-2}\ \text{mol/L·atm}$.

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Q14. What is the freezing point (in °C) of a solution prepared by dissolving 11.3 g of Ca(NO3)2 (formula weight = 164 g/mol) in 115 g of water? The molal freezing-point depression constant for water is 1.86°C/m.

Background

Topic: Colligative Properties – Freezing Point Depression

This question tests your ability to calculate the freezing point depression of a solution using molality and the van't Hoff factor.

Key Terms and Formulas:

• Freezing Point Depression: $\Delta T_f = i K_f m$

• $i$ = van't Hoff factor (number of particles the solute dissociates into)

• $K_f$ = freezing point depression constant (°C/m)

• $m$ = molality (mol solute/kg solvent)

Step-by-Step Guidance

1. Calculate moles of Ca(NO3)2: $n = \frac{11.3\ \text{g}}{164\ \text{g/mol}}$.

2. Calculate molality: $m = \frac{\text{mol solute}}{0.115\ \text{kg solvent}}$.

3. Determine the van't Hoff factor $i$ for Ca(NO3)2 (number of ions produced per formula unit).

4. Calculate $\Delta T_f = i K_f m$.

5. Subtract $\Delta T_f$ from the normal freezing point of water (0°C) to find the new freezing point.

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Q15. The cooling system of an automobile is filled with a solution formed by mixing 4.71 L of water (density = 1.00 g/mL) and 0.290 L of ethylene glycol, C2H6O2 (density = 1.12 g/mL). What is the freezing point of the mixture in Celsius?

Background

Topic: Colligative Properties – Freezing Point Depression

This question tests your ability to calculate the freezing point depression for a solution using the masses and densities of the components.

Key Terms and Formulas:

• Freezing Point Depression: $\Delta T_f = i K_f m$

• $i$ for ethylene glycol is 1 (nonelectrolyte)

• $m$ = molality = mol solute / kg solvent

• Density: $\text{mass} = \text{volume} \times \text{density}$

Step-by-Step Guidance

1. Calculate the mass of water and ethylene glycol using their volumes and densities.

2. Convert the mass of ethylene glycol to moles (molar mass = 62.07 g/mol).

3. Calculate the molality: $m = \frac{\text{mol solute}}{\text{kg solvent}}$.

4. Use $\Delta T_f = i K_f m$ to find the freezing point depression.

5. Subtract $\Delta T_f$ from 0°C to find the new freezing point.

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Q16. What is the osmotic pressure of a solution in atm formed by dissolving 89.5 mg of aspirin (C9H8O4) in 0.500 L of water at 25°C?

Background

Topic: Colligative Properties – Osmotic Pressure

This question tests your ability to calculate the osmotic pressure of a solution using the ideal gas law for solutions.

Key Terms and Formulas:

• Osmotic Pressure: $\Pi = MRT$

• $M$ = molarity (mol/L)

• $R$ = 0.0821 L·atm·mol-1·K-1

• $T$ = temperature in Kelvin

Step-by-Step Guidance

1. Convert 89.5 mg aspirin to grams, then to moles (molar mass = 180.16 g/mol).

2. Calculate the molarity: $M = \frac{\text{mol solute}}{0.500\ \text{L}}$.

3. Convert 25°C to Kelvin: $T = 25 + 273.15$.

4. Plug values into the osmotic pressure formula: $\Pi = MRT$.

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Q17. What is the limiting value of the van’t Hoff factor for Na2CO3?

Background

Topic: Colligative Properties – van’t Hoff Factor

This question tests your understanding of the van’t Hoff factor, which is the number of particles a compound produces in solution.

Key Terms and Concepts:

• van’t Hoff Factor ($i$): Number of ions a formula unit dissociates into in solution.

• Na2CO3 dissociates into 2 Na+ and 1 CO32-.

Step-by-Step Guidance

1. Write the dissociation equation for Na2CO3 in water.

2. Count the total number of ions produced per formula unit.

3. The limiting value of $i$ is the total number of ions formed.

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