Exam 3 – Chapter 2 8, 9 – Gases And Solutions Name: CHM 101

Exam 3 – chap’2 8, 9 – gases and solutions Name: CHM 101

Identify the gas. a. KCl b. NaCl c. NH3 d. NH4Cl

What is the minimum temperature needed to dissolve 40.0 g of NH4Cl in 100 g of H₂O? a. 25°C b. 30°C c. 37°C d. 40°C

What is the maximum mass of sodium nitrate that may be dissolved in 50.0 g of water at 37.0°C? a. 32 g b. 37 g c. 52 g d. 87 g

A student saturates 100 g of water with NaNO₃ at 55°C then cools it to 40°C. What mass will be precipitated? a. 14 g b. 22 g c. 105 g d. 119 g

The following are solutions of soluble compounds at 0.10 M concentration. Which solution contains the greatest number of ions? a. 0.10 M NaCl b. 0.10 M FeCl₃ c. 0.10 M CaCl₂ d. 0.10 M KBr

Which species is the precipitate if ammonium sulfate, (NH₄)₂SO₄, is mixed with barium hydroxide, Ba(OH)₂? a. BaSO₄ b. NH₄OH c. Ba₂SO₄ d. all are soluble

Part II Brief Response

Write the letter of the best response to each question in the table. (30 pt)

For an ideal gas under constant volume, if the pressure is tripled, what happens to the Kelvin temperature? (Answer: The temperature triples, due to Gay-Lussac’s law)

What is the temperature at which all molecular motion is a minimum? (Answer: Absolute zero, 0 K)

Which term is the force per unit area or collisions with the walls of the container? (Answer: Pressure)

A flask contains a mixture of nitrogen and oxygen gases at STP. If the partial pressure of the nitrogen is 40.0 kPa, then what is the partial pressure of the oxygen? (Answer: 100 kPa total pressure - 40.0 kPa = 60.0 kPa)

A flask contains 1.00 mole of neon gas and 2.00 moles of argon gas. If the total pressure of the gases is 5.0 atm, then what is the partial pressure of the argon gas? (Answer: Partial pressure of argon = (2/3) × 5.0 atm = 3.33 atm)

Rank the following gases high to low density: CH₄, CO, He, O₂, SO₂. (Answer: He

Alex cooled a gas from 50.0°C to 25.0°C at constant volume. Which of the following decreased? (Answer: III. The speed of the molecules, as it depends on temperature)

What volume of O₂(g) is required to react with excess CS₂(l) to produce 4.0 L of CO₂(g) at STP? (Answer: According to stoichiometry, 3 L O₂ produces 1 L CO₂, so 4.0 L CO₂ requires 12.0 L O₂)

Jimmy fills a teakettle with water, and begins to heat. When the water boils, why does the kettle whistle? (Answer: Steam escapes through a narrow opening, producing sound)

True/False corrections:

- The distance between gas particles is very large compared with the size of the particles. (True)

- Gases diffuse from areas of high pressure to areas of low pressure. (True)

- Gases have vibrational, rotational, and translational motion. (True)

- The pressure and volume of a gas are directly related when temperature is constant. (False; they are inversely related, per Boyle’s Law)

In which flask do the particles have the highest average speed? (Answer: He, because lighter gases move faster at the same temperature)

Which flask contains the greatest number of particles? (Answer: All contain the same number of particles, if temperature, volume, and moles are same, but since all are 1 L at same T and P, they contain the same number of particles)

Constructed Response

19. a. Sodium carbonate (Na₂CO₃) is soluble in water. (Based on solubility rules, sodium salts are generally soluble.)

b. When sodium carbonate dissolves in water, the aqueous solution contains Na⁺, CO₃²⁻ ions, and possibly ions from impurities. The chemical formula present in water is Na₂CO₃ (aq).

20. The net ionic equation for the reaction between potassium sulfate and calcium hydroxide is:

K₂SO₄ (aq) + Ca(OH)₂ (aq) → CaSO₄ (s) + 2 KOH (aq)

Net ionic: Ca²⁺ (aq) + SO₄²⁻ (aq) → CaSO₄ (s)

21. The molarity of magnesium acetate in solution:

Mass of Mg(HCO₃)₂ = 4.87 g. Molar mass of Mg(HCO₃)₂ ≈ 146.36 g/mol.

Number of moles = 4.87 g / 146.36 g/mol ≈ 0.0333 mol.

Concentration (M) = moles / volume (L) = 0.0333 mol / 0.100 L = 0.333 M.

22. Adding 29.29 mL of 4.27 M solution into water to reach a total volume of 172 mL:

Total moles of solute added = 4.27 mol/L × 0.02929 L ≈ 0.125 mol.

Final molarity = moles / volume (L) = 0.125 mol / 0.172 L ≈ 0.727 M.

23. Moles of BaCl₂ needed for a 0.576 M solution in 2.75 L:

moles = molarity × volume = 0.576 mol/L × 2.75 L ≈ 1.584 mol.

24. Magnesium sulfate (MgSO₄) dissolved in water:

- Moles of solute: 26.0 g / 120.37 g/mol ≈ 0.216 mol.

- Molality = moles / kg solvent = 0.216 mol / 1.25 kg ≈ 0.173 mol/kg.

- Freezing point depression: ΔTf = i × Kf × molality = 2 × 1.86 °C·kg/mol × 0.173 mol/kg ≈ 0.644 °C.

- New boiling point (assuming normal boiling point is 100°C): 100°C + ΔTb (note: problem asks for boiling point change; actual calculation):

- ΔTb = i × Kb × molality = 2 × 0.52 °C·kg/mol × 0.173 mol/kg ≈ 0.18°C. (Since boiling point elevation is usually calculated, but here, focus is on depression and change; the actual change is approximately 0.18°C.)

Thus, the solution’s boiling point increases slightly, approximately to 100.18°C, considering experimental factors and the calculated ΔTb.

References

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