Hard MCAT Molarity Practice Questions

Mastering concentration calculations is a non-negotiable skill for any pre-medical student aiming for a competitive score on the Chemical and Physical Foundations of Biological Systems section. These Hard MCAT Molarity Practice Questions are designed to push your understanding beyond simple division, challenging you with multi-step dilutions, density conversions, and complex stoichiometry. By integrating these problems into your retrieval practice study plan, you can ensure that these mathematical maneuvers become second nature on test day.

Concept Explanation

Molarity is the measure of solute concentration in a solution, defined specifically as the number of moles of solute per liter of total solution.

The fundamental formula for molarity is $M = \frac{n}{V}$, where $M$ is molarity (mol/L), $n$ is the amount of solute in moles, and $V$ is the volume of the solution in liters. While the basic definition is straightforward, the MCAT rarely presents problems in a vacuum. You will often need to convert between units, such as grams to moles using molar mass, or milliliters to liters. Furthermore, you must distinguish between molarity (moles/Liter) and molality (moles/kg solvent), as well as normality (equivalents/Liter), particularly in acid-base chemistry.

On the MCAT, molarity problems frequently appear in the context of colligative properties, titration, and reaction kinetics. Harder questions often involve density (g/mL) or mass percent (% w/w) which require you to find the mass of the solution before finding the volume. Additionally, the dilution equation $M_1V_1 = M_2V_2$ is a staple of laboratory-based passages. Understanding that molarity is temperature-dependent (because volume changes with temperature) while molality is not is a common conceptual trap used by the AAMC.

Solved Examples

Example 1: Density and Mass Percent
A concentrated solution of $H_2SO_4$ is 98% by mass and has a density of 1.84 g/mL. Calculate the molarity of this solution. (Molar mass of $H_2SO_4 = 98.08 \text{ g/mol}$).

1. Assume a basis of 1 Liter (1000 mL) of solution to simplify calculations.
2. Calculate the total mass of 1 L of solution: $$\text{Mass} = 1000 \text{ mL} \times 1.84 \text{ g/mL} = 1840 \text{ g}$$
3. Determine the mass of the solute ($H_2SO_4$) based on the 98% mass percentage: $$1840 \text{ g solution} \times 0.98 = 1803.2 \text{ g } H_2SO_4$$
4. Convert the mass of solute to moles: $$n = \frac{1803.2 \text{ g}}{98.08 \text{ g/mol}} \approx 18.39 \text{ moles}$$
5. Since we assumed a volume of 1 L, the molarity is $18.39 \text{ M}$.

Example 2: Complex Dilution
You have 500 mL of a 0.2 M $NaCl$ solution. You add 300 mL of a 0.5 M $NaCl$ solution and then dilute the final mixture with water to a total volume of 2.0 L. What is the final molarity?

1. Calculate moles from the first solution: $$n_1 = 0.5 \text{ L} \times 0.2 \text{ mol/L} = 0.1 \text{ moles}$$
2. Calculate moles from the second solution: $$n_2 = 0.3 \text{ L} \times 0.5 \text{ mol/L} = 0.15 \text{ moles}$$
3. Find the total moles of solute: $$n_{total} = 0.1 + 0.15 = 0.25 \text{ moles}$$
4. Divide the total moles by the final volume (2.0 L): $$M_{final} = \frac{0.25 \text{ moles}}{2.0 \text{ L}} = 0.125 \text{ M}$$

Example 3: Stoichiometry and Molarity
How many milliliters of 0.15 M $NaOH$ are required to completely neutralize 50 mL of 0.1 M $H_2SO_4$?

1. Write the balanced equation: $$H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$$
2. Calculate moles of $H_2SO_4$: $$n = 0.050 \text{ L} \times 0.1 \text{ mol/L} = 0.005 \text{ moles}$$
3. Use the stoichiometric ratio (2 moles $NaOH$ per 1 mole $H_2SO_4$): $$0.005 \text{ moles } H_2SO_4 \times 2 = 0.01 \text{ moles } NaOH$$
4. Solve for the volume of $NaOH$: $$V = \frac{n}{M} = \frac{0.01 \text{ moles}}{0.15 \text{ M}} \approx 0.0667 \text{ L} = 66.7 \text{ mL}$$

Practice Questions

1. A biological buffer is prepared by dissolving 12.1 g of Tris base (molar mass = 121.1 g/mol) in enough water to make 250 mL of solution. A 10 mL aliquot of this solution is then diluted to a final volume of 500 mL. What is the molarity of the final diluted solution?

2. A sample of seawater has a chloride ion ($Cl^-$) concentration of 0.55 M. If the density of the seawater is 1.025 g/mL, what is the concentration of chloride in parts per million (ppm)? (Atomic mass of $Cl = 35.45 \text{ g/mol}$).

3. Calculate the molarity of a 30% (w/w) aqueous solution of hydrogen peroxide ($H_2O_2$) given that the solution density is 1.11 g/mL. (Molar mass of $H_2O_2 = 34.01 \text{ g/mol}$).

4. A student mixes 150 mL of 0.4 M $CaCl_2$ with 250 mL of 0.2 M $AgNO_3$. Assuming the reaction goes to completion and $AgCl$ precipitates out, what is the final molarity of chloride ions remaining in the solution? (Assume volumes are additive).

5. An unknown diprotic acid ($H_2A$) weighing 0.600 g requires 30.0 mL of 0.200 M $NaOH$ to reach the second equivalence point. What is the molar mass of the acid?

6. If the solubility of $CaF_2$ in water is $2.1 \times 10^{-4} \text{ M}$ at 25°C, what is the molarity of fluoride ions in a saturated solution?

7. A stock solution of 12.0 M $HCl$ is used to prepare 2.5 L of a 0.10 M $HCl$ solution. However, the student accidentally uses a 2.0 L volumetric flask instead of a 2.5 L flask but still adds the amount of stock solution calculated for 2.5 L. What is the resulting molarity?

8. A 5.0 mL sample of gastric juice ($HCl$) is titrated with 0.01 M $NaOH$. If it takes 20.0 mL of $NaOH$ to neutralize the sample, what is the pH of the gastric juice? (Assume $HCl$ is a strong acid and fully dissociates).

9. A solution is prepared by mixing 100 mL of 0.5 M glucose and 400 mL of 0.1 M sucrose. What is the total molarity of sugar molecules in the final solution?

10. How many grams of $Na_2SO_4$ (molar mass = 142 g/mol) are required to prepare 500 mL of a solution where the sodium ion ($Na^+$) concentration is 0.4 M?

Answers & Explanations

1. Answer: 0.008 M
First, find the initial molarity: $n = \frac{12.1 \text{ g}}{121.1 \text{ g/mol}} = 0.1 \text{ moles}$. Initial $M = \frac{0.1 \text{ mol}}{0.250 \text{ L}} = 0.4 \text{ M}$. Next, use the dilution equation $M_1V_1 = M_2V_2$: $(0.4 \text{ M})(10 \text{ mL}) = (M_2)(500 \text{ mL})$. Solving for $M_2$: $M_2 = \frac{4}{500} = 0.008 \text{ M}$.

2. Answer: 19,026 ppm
PPM is defined as (mg solute / kg solution). In 1 L of seawater: $\text{Mass} = 1000 \text{ mL} \times 1.025 \text{ g/mL} = 1025 \text{ g} = 1.025 \text{ kg}$. Moles of $Cl^- = 0.55 \text{ mol}$. Mass of $Cl^- = 0.55 \text{ mol} \times 35.45 \text{ g/mol} = 19.4975 \text{ g} = 19,497.5 \text{ mg}$. $\text{PPM} = \frac{19,497.5 \text{ mg}}{1.025 \text{ kg}} \approx 19,022 \text{ ppm}$. (Slight variation due to rounding).

3. Answer: 9.79 M
Assume 1 L of solution. Total mass = 1110 g. Mass of $H_2O_2 = 1110 \times 0.30 = 333 \text{ g}$. Moles of $H_2O_2 = \frac{333 \text{ g}}{34.01 \text{ g/mol}} = 9.79 \text{ mol}$. Since volume is 1 L, Molarity = 9.79 M.

4. Answer: 0.175 M
Initial moles $Cl^- = 0.150 \text{ L} \times 0.4 \text{ M} \times 2 = 0.12 \text{ moles}$. Initial moles $Ag^+ = 0.250 \text{ L} \times 0.2 \text{ M} = 0.05 \text{ moles}$. Reaction: $Ag^+ + Cl^- \rightarrow AgCl(s)$. 0.05 moles of $Ag^+$ react with 0.05 moles of $Cl^-$. Remaining moles $Cl^- = 0.12 - 0.05 = 0.07 \text{ moles}$. Total volume = 400 mL = 0.4 L. Final $[Cl^-] = \frac{0.07}{0.4} = 0.175 \text{ M}$.

5. Answer: 200 g/mol
Moles of $NaOH = 0.030 \text{ L} \times 0.200 \text{ M} = 0.006 \text{ moles}$. Since the acid is diprotic, 2 moles of $NaOH$ react with 1 mole of acid. Moles of acid = $\frac{0.006}{2} = 0.003 \text{ moles}$. Molar mass = $\frac{0.600 \text{ g}}{0.003 \text{ mol}} = 200 \text{ g/mol}$.

6. Answer: $4.2 \times 10^{-4} \text{ M}$
The dissociation of calcium fluoride is $CaF_2 \rightarrow Ca^{2+} + 2F^-$. If the solubility is $s$, the concentration of fluoride ions is $2s$. Thus, $[F^-] = 2 \times (2.1 \times 10^{-4} \text{ M}) = 4.2 \times 10^{-4} \text{ M}$.

7. Answer: 0.125 M
First, calculate the volume of stock needed for the intended 2.5 L: $V_{stock} = \frac{0.1 \times 2.5}{12.0} = 0.02083 \text{ L}$. This amount of solute (0.25 moles) was actually placed in 2.0 L. $M_{actual} = \frac{0.25 \text{ moles}}{2.0 \text{ L}} = 0.125 \text{ M}$.

8. Answer: 1.4
Moles $NaOH = 0.020 \text{ L} \times 0.01 \text{ M} = 0.0002 \text{ moles}$. Since the ratio is 1:1, moles $HCl = 0.0002$. Concentration of $HCl = \frac{0.0002 \text{ mol}}{0.005 \text{ L}} = 0.04 \text{ M}$. $pH = -\log[H^+] = -\log(0.04)$. Since $-\log(0.01) = 2$ and $-\log(0.1) = 1$, the value is between 1 and 2. Specifically, $-\log(4 \times 10^{-2}) = 2 - \log(4) \approx 2 - 0.6 = 1.4$.

9. Answer: 0.18 M
Moles glucose = $0.1 \text{ L} \times 0.5 \text{ M} = 0.05 \text{ moles}$. Moles sucrose = $0.4 \text{ L} \times 0.1 \text{ M} = 0.04 \text{ moles}$. Total sugar moles = 0.09 moles. Total volume = 0.5 L. $M = \frac{0.09}{0.5} = 0.18 \text{ M}$.

10. Answer: 14.2 g
Desired $[Na^+] = 0.4 \text{ M}$. Since each $Na_2SO_4$ provides 2 sodium ions, the required molarity of $Na_2SO_4$ is 0.2 M. Moles of $Na_2SO_4 = 0.5 \text{ L} \times 0.2 \text{ M} = 0.1 \text{ moles}$. Mass = $0.1 \text{ moles} \times 142 \text{ g/mol} = 14.2 \text{ g}$.

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

• AAdding more solvent to the solution
• BAllowing some of the solvent to evaporate
• CIncreasing the temperature of the solution
• DAdding more of a different, non-reacting solute

Frequently Asked Questions

What is the difference between molarity and molality?

Molarity is the number of moles of solute per liter of solution, whereas molality is the number of moles of solute per kilogram of solvent. Molarity is affected by temperature due to volume changes, while molality remains constant regardless of temperature.

How do you convert mass percent to molarity?

To convert mass percent to molarity, multiply the mass percent (as a decimal) by the density of the solution (in g/L) to find the mass of solute per liter. Then, divide this mass by the molar mass of the solute to obtain the molarity.

Does adding solute always change the molarity?

Yes, adding more of the same solute will increase the molarity because the number of moles increases more significantly than the negligible volume change of the solid solute. If you add a different solute, the molarity of the original solute remains the same, but the total particle concentration increases.

Why does the MCAT use normality instead of molarity sometimes?

Normality is used primarily in acid-base and redox chemistry to account for the number of reactive equivalents per liter. For example, a 1 M solution of $H_2SO_4$ is 2 N because it provides two moles of hydrogen ions per mole of acid.

What is a common mistake on hard MCAT molarity questions?

A frequent error is failing to account for the dissociation of ionic compounds, such as forgetting that 1 M $CaCl_2$ produces 2 M of chloride ions. Additionally, students often forget to convert milliliters to liters when using the standard molarity formula.

For more advanced science strategies, check out our guide on retrieval practice for STEM subjects or learn how to transform your medical education through active recall.