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    MCAT Molarity Practice Questions with Answers

    May 9, 202610 min read46 views
    MCAT Molarity Practice Questions with Answers

    MCAT Molarity Practice Questions with Answers

    Mastering MCAT Molarity involves understanding the relationship between solute amount and solution volume, a fundamental skill for the Chemical and Physical Foundations of Biological Systems section. Whether you are calculating the concentration of a physiological saline solution or determining the stoichiometric requirements of a titration, molarity is the most common unit of concentration you will encounter on exam day. By integrating these concepts into your retrieval practice for medical students, you can ensure that these formulas become second nature.

    Concept Explanation

    Molarity (M) is defined as the number of moles of solute dissolved per liter of solution. It is a measure of concentration that relates the amount of substance to the total volume of the mixture. The standard formula for molarity is:

    M = n V M = \frac{n}{V}

    In this equation, M M represents molarity in moles per liter (mol/L), n n is the number of moles of solute, and V V is the total volume of the solution in liters (L). It is vital to remember that the volume in the denominator is the volume of the entire solution, not just the solvent. Because volume is temperature-dependent due to thermal expansion, molarity can change slightly with temperature, a detail occasionally tested in conceptual MCAT questions.

    On the MCAT, you will often need to convert between grams and moles using the molar mass before applying the molarity formula. Additionally, dilution problems are a staple of the exam, utilizing the conservation of moles:

    M 1 V 1 = M 2 V 2 M_1V_1 = M_2V_2

    Using retrieval practice for STEM subjects can help you quickly recall these units and conversion factors under the pressure of a timed exam. High-authority resources like LibreTexts Chemistry provide excellent theoretical backgrounds on these aqueous solution properties.

    Solved Examples

    Example 1: Basic Molarity Calculation
    Calculate the molarity of a solution prepared by dissolving 58.5 grams of N a C l NaCl (molar mass = 58.5 g/mol) in enough water to make 500 mL of solution.

    1. Convert mass to moles: n = 58.5  g 58.5  g/mol = 1.0  mole n = \frac{58.5 \text{ g}}{58.5 \text{ g/mol}} = 1.0 \text{ mole} .
    2. Convert volume to liters: 500  mL = 0.5  L 500 \text{ mL} = 0.5 \text{ L} .
    3. Calculate molarity: M = 1.0  mol 0.5  L = 2.0  M M = \frac{1.0 \text{ mol}}{0.5 \text{ L}} = 2.0 \text{ M} .

    Example 2: Dilution Logic
    A student has 100 mL of a 6.0 M H C l HCl stock solution. If they dilute this to a final volume of 1.0 L, what is the new concentration?

    1. Identify variables: M 1 = 6.0  M M_1 = 6.0 \text{ M} , V 1 = 0.1  L V_1 = 0.1 \text{ L} , V 2 = 1.0  L V_2 = 1.0 \text{ L} .
    2. Apply the dilution formula: ( 6.0  M ) ( 0.1  L ) = ( M 2 ) ( 1.0  L ) (6.0 \text{ M})(0.1 \text{ L}) = (M_2)(1.0 \text{ L}) .
    3. Solve for M 2 M_2 : M 2 = 0.6  M M_2 = 0.6 \text{ M} .

    Example 3: Finding Mass from Molarity
    How many grams of g l u c o s e glucose ( C 6 H 12 O 6 C_6H_{12}O_6 , molar mass = 180 g/mol) are needed to prepare 250 mL of a 0.5 M solution?

    1. Convert volume to liters: 0.25  L 0.25 \text{ L} .
    2. Calculate moles needed: n = M × V = 0.5  mol/L × 0.25  L = 0.125  moles n = M \times V = 0.5 \text{ mol/L} \times 0.25 \text{ L} = 0.125 \text{ moles} .
    3. Convert moles to grams: 0.125  mol × 180  g/mol = 22.5  grams 0.125 \text{ mol} \times 180 \text{ g/mol} = 22.5 \text{ grams} .

    Practice Questions

    1. A 2.0 L solution contains 0.4 moles of N a O H NaOH . What is the molarity of the solution?

    2. How many moles of K C l KCl are present in 150 mL of a 0.2 M solution?

    3. A chemist dissolves 40 grams of N a O H NaOH (molar mass = 40 g/mol) in water to create a 2.0 M solution. What is the final volume of the solution in milliliters?

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    4. To what volume should 50 mL of 12 M H 2 S O 4 H_2SO_4 be diluted to prepare a 3.0 M solution?

    5. What is the molarity of a solution made by dissolving 10 grams of M g C l 2 MgCl_2 (molar mass = 95 g/mol) in 200 mL of water? (Assume the volume of the solution is 200 mL).

    6. If 500 mL of 0.1 M C a C l 2 CaCl_2 is mixed with 500 mL of water, what is the final concentration of chloride ions ( C l − Cl^- )?

    7. A biological buffer requires a concentration of 0.15 M N a C l NaCl . How much water must be added to 50 mL of a 1.5 M N a C l NaCl stock to reach this concentration?

    8. Calculate the molarity of a 20% (w/v) solution of N a O H NaOH . (w/v means 20g of solute in 100mL of solution).

    9. A sample of 0.5 M N a 3 P O 4 Na_3PO_4 has a volume of 2.0 L. How many moles of sodium ions ( N a + Na^+ ) are in the sample?

    10. What is the density of a 5.0 M solution of ethanol ( C 2 H 5 O H C_2H_5OH , molar mass = 46 g/mol) if the solution has a total mass of 1200 g and a volume of 1.0 L?

    Answers & Explanations

    1. Answer: 0.2 M
    Using the formula M = n / V M = n / V , we have 0.4  moles / 2.0  L = 0.2  M 0.4 \text{ moles} / 2.0 \text{ L} = 0.2 \text{ M} . This is a straightforward application of the definition.

    2. Answer: 0.03 moles
    Rearrange the formula to n = M × V n = M \times V . Convert 150 mL to 0.15 L. Then, 0.2  mol/L × 0.15  L = 0.03  moles 0.2 \text{ mol/L} \times 0.15 \text{ L} = 0.03 \text{ moles} .

    3. Answer: 500 mL
    First, find the moles: 40  g / 40  g/mol = 1.0  mole 40 \text{ g} / 40 \text{ g/mol} = 1.0 \text{ mole} . Then, use V = n / M V = n / M : 1.0  mol / 2.0  M = 0.5  L 1.0 \text{ mol} / 2.0 \text{ M} = 0.5 \text{ L} . Convert to mL: 500  mL 500 \text{ mL} .

    4. Answer: 200 mL
    Using M 1 V 1 = M 2 V 2 M_1V_1 = M_2V_2 : ( 12  M ) ( 50  mL ) = ( 3.0  M ) ( V 2 ) (12 \text{ M})(50 \text{ mL}) = (3.0 \text{ M})(V_2) . Solving for V 2 V_2 gives 600 / 3.0 = 200  mL 600 / 3.0 = 200 \text{ mL} .

    5. Answer: 0.53 M
    Moles of M g C l 2 = 10  g / 95  g/mol ≈ 0.105  moles MgCl_2 = 10 \text{ g} / 95 \text{ g/mol} \approx 0.105 \text{ moles} . Volume is 0.2 L. M = 0.105 / 0.2 = 0.526  M M = 0.105 / 0.2 = 0.526 \text{ M} , rounded to 0.53 M.

    6. Answer: 0.1 M
    The solution is diluted 1:1, so the concentration of C a C l 2 CaCl_2 becomes 0.05 M. Since each C a C l 2 CaCl_2 dissociates into two C l − Cl^- ions, the concentration of C l − Cl^- is 2 × 0.05  M = 0.1  M 2 \times 0.05 \text{ M} = 0.1 \text{ M} .

    7. Answer: 450 mL
    Using M 1 V 1 = M 2 V 2 M_1V_1 = M_2V_2 : ( 1.5 ) ( 50 ) = ( 0.15 ) ( V 2 ) (1.5)(50) = (0.15)(V_2) , so V 2 = 500  mL V_2 = 500 \text{ mL} . The question asks how much water must be added, so 500  mL − 50  mL = 450  mL 500 \text{ mL} - 50 \text{ mL} = 450 \text{ mL} .

    8. Answer: 5.0 M
    20% w/v means 20 g in 100 mL, which is 200 g in 1.0 L. Moles of N a O H = 200  g / 40  g/mol = 5.0  moles NaOH = 200 \text{ g} / 40 \text{ g/mol} = 5.0 \text{ moles} . Since this is in 1.0 L, the molarity is 5.0 M.

    9. Answer: 3.0 moles
    Total moles of N a 3 P O 4 = 0.5  M × 2.0  L = 1.0  mole Na_3PO_4 = 0.5 \text{ M} \times 2.0 \text{ L} = 1.0 \text{ mole} . Each mole of N a 3 P O 4 Na_3PO_4 contains 3 moles of N a + Na^+ , so 1.0 × 3 = 3.0  moles 1.0 \times 3 = 3.0 \text{ moles} .

    10. Answer: 1.2 g/mL
    Density is total mass / total volume. 1200  g / 1000  mL = 1.2  g/mL 1200 \text{ g} / 1000 \text{ mL} = 1.2 \text{ g/mL} . Note that the molarity was extra information not needed for the density calculation itself, a common MCAT distractor.

    Interactive quizQuestion 1 of 5

    1. Which of the following changes would increase the molarity of a solution?

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    Frequently Asked Questions

    What is the difference between molarity and molality?

    Molarity is the moles of solute per liter of solution, whereas molality is the moles of solute per kilogram of solvent. Molarity is volume-dependent and temperature-sensitive, while molality remains constant regardless of temperature or pressure changes.

    How do you convert from molarity to mass?

    Multiply the molarity by the volume of the solution in liters to find the number of moles, then multiply the result by the molar mass of the solute. This multi-step process is a frequent requirement in stoichiometry problems on the MCAT.

    Does adding solute increase the volume of a solution?

    In most practical laboratory scenarios, adding a significant amount of solid solute will slightly increase the total volume of the solution. This is why standard procedure involves dissolving the solute in a small amount of solvent first and then "diluting to the mark" in a volumetric flask.

    How does ion dissociation affect molarity calculations?

    When calculating the molarity of specific ions, you must multiply the molarity of the parent compound by the stoichiometric coefficient of the ion in the chemical formula. For example, a 1 M solution of M g C l 2 MgCl_2 is 2 M in chloride ions because each unit releases two C l − Cl^- ions.

    What are common units for volume on the MCAT?

    The MCAT frequently uses liters (L), milliliters (mL), and cubic centimeters (cm³ or cc), where 1 mL is exactly equal to 1 cm³. Always ensure your volume is in liters before plugging it into the M = n / V M = n / V formula to avoid power-of-ten errors.

    For more strategies on how to handle difficult science passages, check out our guide on retrieval practice vs practice tests to optimize your study schedule.

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    Michael Danquah, MS, PhD

    Reviewed by

    Michael Danquah, MS, PhD

    Dr. Michael Danquah is a professor of pharmaceutical sciences and founder of several educational technology platforms focused on improving student learning and performance.

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