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:

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

In this equation, $M$ represents molarity in moles per liter (mol/L), $n$ is the number of moles of solute, and $V$ 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:

$$M_1V_1 = M_2V_2$$

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 $NaCl$ (molar mass = 58.5 g/mol) in enough water to make 500 mL of solution.

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

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

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

Example 3: Finding Mass from Molarity
How many grams of $glucose$ ($C_6H_{12}O_6$, molar mass = 180 g/mol) are needed to prepare 250 mL of a 0.5 M solution?

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

Practice Questions

1. A 2.0 L solution contains 0.4 moles of $NaOH$. What is the molarity of the solution?

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

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

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

5. What is the molarity of a solution made by dissolving 10 grams of $MgCl_2$ (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 $CaCl_2$ is mixed with 500 mL of water, what is the final concentration of chloride ions ($Cl^-$)?

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

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

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

10. What is the density of a 5.0 M solution of ethanol ($C_2H_5OH$, 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 $M = n / V$, we have $0.4 \text{ moles} / 2.0 \text{ L} = 0.2 \text{ M}$. This is a straightforward application of the definition.

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

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

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

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

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

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

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

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

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

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 $MgCl_2$ is 2 M in chloride ions because each unit releases two $Cl^-$ 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 $M = n / V$ 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.