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MCAT · Chemical and Physical Foundations · General Chemistry
Stoichiometry and solutions is part of General Chemistry in the MCAT Chemical and Physical Foundations section. The questions below are written in the style of the real test, each with the answer and a worked explanation.
Question types you will see here include limiting reagent and percent yield, molarity and dilution, ksp and molar solubility and order of precipitation.
Common traps: choosing the smaller mass rather than the smaller mole-to-coefficient ratio; ignoring the balanced coefficients; inverting the percent yield ratio.
Question 1
40 g of a solute with molar mass 40 g/mol is dissolved in water to make 2 L of solution. What is the molarity?
0.25 M
0.5 M
2 M
20 M
Answer: B. 0.5 M
Moles = 40/40 = 1 mol, and molarity = mol/L = 1/2 = 0.5 M. Multiplying by volume instead of dividing gives 2.
Question 2
What is the concentration of Na⁺ in a 0.3 M solution of Na₂SO₄, assuming complete dissociation?
0.06 M
0.15 M
0.3 M
0.6 M
Answer: D. 0.6 M
Each formula unit of Na₂SO₄ releases 2 Na⁺, so [Na⁺] = 2 × 0.3 M = 0.6 M. The salt's own molarity (0.3 M) counts formula units, not ions.
Question 3
What is the concentration of Cl⁻ in a 0.25 M solution of MgCl₂, assuming complete dissociation?
0.5 M
0.75 M
1 M
5 M
Answer: A. 0.5 M
Each formula unit of MgCl₂ releases 2 Cl⁻, so [Cl⁻] = 2 × 0.25 M = 0.5 M. The salt's own molarity (0.25 M) counts formula units, not ions.
Question 4
A sparingly soluble salt MX2 has Ksp = 3.2 × 10⁻¹¹. Its molar solubility in pure water is closest to:
5.7 × 10⁻⁶ M
2.0 × 10⁻⁴ M
3.2 × 10⁻⁴ M
4.0 × 10⁻⁴ M
Answer: B. 2.0 × 10⁻⁴ M
If s mol/L dissolves, [M²⁺] = s and [X⁻] = 2s, so Ksp = s(2s)² = 4s³. Then s³ = (3.2 × 10⁻¹¹)/4 = 8.0 × 10⁻¹², and s = 2.0 × 10⁻⁴ M. The 4.0 × 10⁻⁴ M choice is [X⁻], not the solubility of the salt, and 3.2 × 10⁻⁴ M comes from taking the cube root of Ksp without dividing by 4.
Question 5
25.0 mL of 2.0 M NaCl is diluted to 200 mL. Then 60 mL of this solution is mixed with 40 mL of 0.50 M CaCl₂. What is [Cl⁻] in the final 100-mL mixture?
0.15 M
0.35 M
0.40 M
0.55 M
Answer: D. 0.55 M
The diluted NaCl is 0.25 M, contributing 0.25 × 60 = 15 mmol Cl⁻; CaCl₂ gives 2 × 0.50 × 40 = 40 mmol. Total 55 mmol/100 mL = 0.55 M (D). A counts only the NaCl's chloride, C only the CaCl₂'s, and B gives CaCl₂ one chloride per formula unit.
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