AP Chemistry · Topic 3.8
Representations of Solutions Practice
Part of Properties of Substances and Mixtures.(SPQ-3.B)
Practice questions
15
Sample questions
5 of 15 — sign in to practice the rest with adaptive difficulty and mastery tracking.
Sample 1difficulty 1/5
The molarity of the solution is:
- A
0.025 M
- Bcheck_circle
0.10 M
- C
1.0 M
- D
0.25 M
Why
M = mol/L = 0.025 mol / 0.250 L = 0.10 M.
- A
Sample 2difficulty 2/5
"5 ppm of lead" in water means
- A
5% by mass
- B
5 mol per liter
- Ccheck_circle
5 mg of lead per liter (or 5 µg per gram)
- D
5 g of lead per liter
Why
ppm = parts per million. For dilute aqueous solutions: 1 ppm ≈ 1 mg/L.
- A
Sample 3difficulty 2/5
What is the molarity of a solution made by dissolving 0.5 mol of NaCl in 250 mL of water?
- A
1.0 M
- B
0.125 M
- C
0.5 M
- Dcheck_circle
2.0 M
Why
M = 0.5 / 0.250 = 2.0 mol/L.
- A
Sample 4difficulty 2/5
100 mL of 0.10 M HCl is diluted to a final volume of 500 mL.
The new molarity is:
- A
0.10 M
- B
0.50 M
- Ccheck_circle
0.020 M
- D
0.0050 M
Why
M1V1 = M2V2; (0.10)(100) = M2(500), giving M2 = 0.020 M.
- A
Sample 5difficulty 2/5
The dilution equation is
- A
M₁ = M₂
- B
n₁ + n₂ = M
- Ccheck_circle
M₁V₁ = M₂V₂
- D
V₁ = V₂
Why
The number of moles of solute is conserved during dilution: n = M × V is constant.
- A