Medium MCAT Molarity Practice Questions

Mastering MCAT Molarity is a fundamental requirement for the Chemical and Physical Foundations of Biological Systems section of the exam. Molarity, defined as the number of moles of solute per liter of solution, serves as the primary unit of concentration in general chemistry, biochemistry, and physiological calculations. Because the MCAT often combines concentration units with stoichiometry, acid-base chemistry, and kinetics, having a fluid grasp of these conversions is essential for scoring well. Using retrieval practice techniques while solving these medium-difficulty problems will ensure that you can recall the necessary formulas under the pressure of a timed exam.

Concept Explanation

Molarity is the measure of solute concentration in a solution, expressed specifically as the ratio of moles of solute to the total volume of the solution in liters. The mathematical representation is given by the formula:

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

In this equation, $M$ represents molarity (mol/L), $n$ represents the number of moles of solute, and $V$ represents the volume of the solution in liters. When working through Medium MCAT Molarity practice questions, you must often perform preliminary steps, such as converting grams to moles using the molar mass or converting milliliters to liters. It is also important to distinguish molarity from molality (moles per kilogram of solvent) and normality (equivalents per liter), as the AAMC frequently tests your ability to navigate these nuances. Many students find that they make common mistakes by failing to account for the final volume of the solution after multiple reagents are mixed. Always ensure your volume units are in liters before finalizing your calculation.

Solved Examples

Review these worked examples to understand the logical flow required for multi-step molarity problems.

1. Example 1: Calculating Molarity from Mass
   A student dissolves 58.5 grams of $NaCl$ (molar mass = 58.5 g/mol) in enough water to make 500 mL of solution. What is the molarity?
   1. Find the moles of solute: $\frac{58.5 \text{ g}}{58.5 \text{ g/mol}} = 1.0 \text{ mole}$.
   2. Convert volume to liters: $500 \text{ mL} = 0.5 \text{ L}$.
   3. Apply the formula: $M = \frac{1.0 \text{ mol}}{0.5 \text{ L}} = 2.0 \text{ M}$.
2. Example 2: Dilution Calculations
   How much water must be added to 100 mL of a 2.0 M $HCl$ solution to create a 0.5 M solution?
   1. Use the dilution equation: $M_1V_1 = M_2V_2$.
   2. Plug in knowns: $(2.0 \text{ M})(0.1 \text{ L}) = (0.5 \text{ M})(V_2)$.
   3. Solve for $V_2$: $V_2 = \frac{0.2}{0.5} = 0.4 \text{ L}$ (or 400 mL).
   4. Calculate added volume: $400 \text{ mL} - 100 \text{ mL} = 300 \text{ mL}$.
3. Example 3: Ion Concentration
   What is the molar concentration of chloride ions in a 0.15 M solution of $CaCl_2$?
   1. Write the dissociation equation: $CaCl_2 \rightarrow Ca^{2+} + 2Cl^-$.
   2. Identify the ratio: 1 mole of $CaCl_2$ produces 2 moles of $Cl^-$.
   3. Calculate: $0.15 \text{ M} \times 2 = 0.30 \text{ M} \text{ of } Cl^-$.

Practice Questions

Apply your knowledge to the following Medium MCAT Molarity practice questions. Ensure you convert all units to liters and moles before calculating.

1. A 250 mL solution contains 0.05 moles of glucose. If 750 mL of pure water is added to this solution, what is the final molarity of glucose?

2. How many grams of $NaOH$ (molar mass = 40 g/mol) are required to prepare 2.5 L of a 0.5 M solution?

3. A researcher mixes 200 mL of 1.0 M $KCl$ with 300 mL of 2.0 M $KCl$. What is the final molarity of the potassium ions in the mixture?

4. What is the molarity of a 20% (w/v) solution of $MgCl_2$ (molar mass ≈ 95 g/mol)?

5. Calculate the volume (in mL) of 12 M $HCl$ needed to prepare 600 mL of a 0.2 M $HCl$ solution.

6. If the density of a 5.0 M $H_2SO_4$ solution is 1.28 g/mL, what is the mass of 1 liter of this solution?

7. A solution of $Al(NO_3)_3$ has a nitrate ion concentration of 0.9 M. What is the molarity of the aluminum nitrate solute?

8. How many moles of sulfate ions are present in 500 mL of 0.4 M $Fe_2(SO_4)_3$?

9. A biological buffer is prepared by dissolving 12.1 g of Tris base (molar mass = 121 g/mol) in 1 liter of water. What is the molarity?

10. If 50 mL of 0.1 M $AgNO_3$ is mixed with 50 mL of 0.1 M $NaCl$, a precipitate of $AgCl$ forms. Assuming the reaction goes to completion, what is the concentration of nitrate ions in the final supernatant?

Answers & Explanations

1. Answer: 0.05 M
   The initial volume is 250 mL and 750 mL is added, making the final volume 1000 mL (1.0 L). Moles of glucose = 0.05. $M = \frac{0.05 \text{ mol}}{1.0 \text{ L}} = 0.05 \text{ M}$.
2. Answer: 50 g
   First, find the moles: $n = M \times V = 0.5 \text{ M} \times 2.5 \text{ L} = 1.25 \text{ moles}$. Then convert to grams: $1.25 \text{ mol} \times 40 \text{ g/mol} = 50 \text{ g}$.
3. Answer: 1.6 M
   Total moles = $(1.0 \text{ M} \times 0.2 \text{ L}) + (2.0 \text{ M} \times 0.3 \text{ L}) = 0.2 + 0.6 = 0.8 \text{ moles}$. Total volume = $0.2 \text{ L} + 0.3 \text{ L} = 0.5 \text{ L}$. Final $M = \frac{0.8 \text{ mol}}{0.5 \text{ L}} = 1.6 \text{ M}$.
4. Answer: 2.1 M
   20% (w/v) means 20 g of solute in 100 mL of solution. In 1 liter (1000 mL), there would be 200 g. Moles = $\frac{200 \text{ g}}{95 \text{ g/mol}} \approx 2.1 \text{ moles}$. Since this is in 1 L, the molarity is 2.1 M.
5. Answer: 10 mL
   Using $M_1V_1 = M_2V_2$: $(12 \text{ M})(V_1) = (0.2 \text{ M})(600 \text{ mL})$. $12V_1 = 120$. $V_1 = 10 \text{ mL}$.
6. Answer: 1280 g
   Density = mass/volume. For 1 liter (1000 mL): $1.28 \text{ g/mL} \times 1000 \text{ mL} = 1280 \text{ g}$. (Note: This is the mass of the solution, not just the solute).
7. Answer: 0.3 M
   $Al(NO_3)_3$ dissociates into 1 $Al^{3+}$ and 3 $NO_3^-$. If $[NO_3^-] = 0.9 \text{ M}$, then the concentration of the parent compound is $\frac{0.9}{3} = 0.3 \text{ M}$.
8. Answer: 0.6 moles
   Moles of $Fe_2(SO_4)_3 = 0.4 \text{ M} \times 0.5 \text{ L} = 0.2 \text{ moles}$. Each mole of compound has 3 moles of $SO_4^{2-}$. Total sulfate moles = $0.2 \times 3 = 0.6 \text{ moles}$.
9. Answer: 0.1 M
   Moles = $\frac{12.1 \text{ g}}{121 \text{ g/mol}} = 0.1 \text{ moles}$. Volume = 1 L. $M = \frac{0.1 \text{ mol}}{1 \text{ L}} = 0.1 \text{ M}$.
10. Answer: 0.05 M
    Nitrate is a spectator ion. Initial moles of $NO_3^- = 0.1 \text{ M} \times 0.05 \text{ L} = 0.005 \text{ moles}$. The final volume is 100 mL (0.1 L). Final concentration = $\frac{0.005 \text{ mol}}{0.1 \text{ L}} = 0.05 \text{ M}$.

1. Which of the following describes a 1.0 M solution of glucose?

• A1.0 gram of glucose in 1.0 liter of water
• B1.0 mole of glucose in 1.0 liter of solution
• C1.0 mole of glucose in 1.0 kilogram of water
• D1.0 gram of glucose in 1.0 milliliter of solution

Frequently Asked Questions

Why does molarity change with temperature?

Molarity is dependent on the volume of the solution, which expands or contracts as temperature changes. Since the number of moles remains constant while volume changes, the molarity value fluctuates slightly with thermal shifts.

What is the difference between molarity and molality?

Molarity measures moles per liter of solution, whereas molality measures moles per kilogram of solvent. Molality is preferred in thermodynamics because it remains constant regardless of temperature or pressure changes.

How do you convert from percent composition to molarity?

Assume a 100 g or 1 L sample, convert the mass of the solute to moles using molar mass, and then divide by the total volume of the solution in liters. For weight/volume percent, the calculation is more direct as the volume is already implied.

Can molarity be used for gases?

Yes, molarity can be applied to gases by using the Ideal Gas Law to determine the number of moles per unit volume. However, partial pressures are more commonly used in MCAT gas phase problems.

What is a standard solution in chemistry?

A standard solution is a solution with a precisely known molarity, often used in titrations to determine the concentration of an unknown analyte. These are typically prepared using primary standards with high purity and stability.