Medium Molarity Practice Questions

Concept Explanation

Molarity is a measure of the concentration of a solute in a solution, defined as the number of moles of solute per liter of solution. It is the most common unit of concentration used in chemistry and is expressed by the symbol 'M' (molar). To calculate molarity, you must divide the total moles of solute by the total volume of the solution in liters. This concept is fundamental for performing stoichiometry calculations in aqueous environments. Unlike molality, molarity is temperature-dependent because the volume of a liquid can expand or contract as temperature changes. For more advanced applications, you can read about molarity on Wikipedia or explore instructional videos on Khan Academy. Understanding molarity allows chemists to predict how much of a substance is present in a specific volume, which is essential for laboratory safety and experimental accuracy.

Solved Examples

These examples demonstrate the step-by-step process for solving typical medium-level molarity problems, including mass-to-molarity conversions and dilution calculations.

1. Example 1: Calculating Molarity from Mass
   How many molar is a solution prepared by dissolving 45.0 grams of Glucose (C₆H₁₂O₆) in enough water to make 500.0 mL of solution?
   
   1. Identify the molar mass of Glucose: (6 × 12.01) + (12 × 1.01) + (6 × 16.00) = 180.18 g/mol.
   2. Convert grams to moles: 45.0 g / 180.18 g/mol = 0.2497 moles.
   3. Convert milliliters to liters: 500.0 mL / 1000 = 0.5000 L.
   4. Calculate Molarity: 0.2497 moles / 0.5000 L = 0.499 M.
2. Example 2: Finding Required Mass
   How many grams of Sodium Hydroxide (NaOH) are needed to prepare 250.0 mL of a 2.00 M solution?
   
   1. Convert volume to liters: 250.0 mL = 0.250 L.
   2. Calculate moles needed: Molarity × Volume = 2.00 mol/L × 0.250 L = 0.500 moles.
   3. Identify molar mass of NaOH: 22.99 + 16.00 + 1.01 = 40.00 g/mol.
   4. Convert moles to grams: 0.500 moles × 40.00 g/mol = 20.0 grams. (For more on this, see moles to grams practice).
3. Example 3: Dilution Problem
   If you dilute 50.0 mL of a 12.0 M HCl stock solution to a final volume of 1.00 L, what is the new molarity?
   
   1. Use the dilution formula: M₁V₁ = M₂V₂.
   2. Identify variables: M₁ = 12.0 M, V₁ = 0.050 L, V₂ = 1.00 L.
   3. Solve for M₂: (12.0 M × 0.050 L) / 1.00 L = 0.600 M.

Practice Questions

1. What is the molarity of a solution containing 15.8 grams of KMnO₄ dissolved in 2.50 L of solution?
2. Calculate the volume (in mL) of 0.750 M Na₂SO₄ solution that contains 10.0 grams of solute.
3. A chemist needs 0.125 moles of HCl for a reaction. How many milliliters of a 3.00 M HCl solution should be measured out?

4. How many grams of CaCl₂ are required to prepare 750 mL of a 0.50 M solution?
5. To what volume should 25.0 mL of 15.0 M HNO₃ be diluted to prepare a 0.200 M solution?
6. If 100.0 mL of water is added to 200.0 mL of a 0.450 M MgSO₄ solution, what is the final molarity? (Assume volumes are additive).
7. What is the molarity of chloride ions in a 0.150 M solution of AlCl₃?
8. A 50.0 mL sample of 0.100 M AgNO₃ is mixed with 50.0 mL of 0.100 M NaCl. How many moles of AgCl precipitate? (Hint: Check limiting reagent concepts).
9. Calculate the molarity of a solution made by dissolving 2.34 g of NaCl in enough water to make 100.0 mL of solution.
10. How many milliliters of 18.0 M H₂SO₄ are required to prepare 2.00 L of 0.250 M H₂SO₄?

Answers & Explanations

1. 0.0400 M: Molar mass of KMnO₄ = 158.03 g/mol. Moles = 15.8 g / 158.03 g/mol = 0.100 mol. Molarity = 0.100 mol / 2.50 L = 0.0400 M.
2. 93.9 mL: Molar mass of Na₂SO₄ = 142.04 g/mol. Moles = 10.0 g / 142.04 g/mol = 0.0704 mol. Volume = Moles / Molarity = 0.0704 mol / 0.750 M = 0.0939 L = 93.9 mL.
3. 41.7 mL: Volume = Moles / Molarity = 0.125 mol / 3.00 mol/L = 0.04167 L = 41.7 mL.
4. 41.6 grams: Moles = 0.50 M × 0.750 L = 0.375 mol. Molar mass of CaCl₂ = 110.98 g/mol. Mass = 0.375 mol × 110.98 g/mol = 41.6 g.
5. 1.88 L: Use M₁V₁ = M₂V₂. (15.0 M)(0.0250 L) = (0.200 M)(V₂). V₂ = 0.375 / 0.200 = 1.875 L.
6. 0.300 M: Initial moles = 0.200 L × 0.450 M = 0.090 mol. Final volume = 200.0 + 100.0 = 300.0 mL = 0.300 L. New Molarity = 0.090 mol / 0.300 L = 0.300 M.
7. 0.450 M: AlCl₃ dissociates into Al³⁺ and 3Cl⁻. Therefore, [Cl⁻] = 3 × [AlCl₃] = 3 × 0.150 M = 0.450 M.
8. 0.00500 moles: Moles AgNO₃ = 0.050 L × 0.100 M = 0.005 mol. Moles NaCl = 0.050 L × 0.100 M = 0.005 mol. Since the ratio is 1:1, 0.005 moles of AgCl form.
9. 0.400 M: Molar mass NaCl = 58.44 g/mol. Moles = 2.34 g / 58.44 g/mol = 0.0400 mol. Molarity = 0.0400 mol / 0.100 L = 0.400 M.
10. 27.8 mL: (18.0 M)(V₁) = (0.250 M)(2.00 L). V₁ = 0.500 / 18.0 = 0.02778 L = 27.8 mL.

Quick Quiz

Frequently Asked Questions

What is the difference between molarity and molality?

Molarity measures moles of solute per liter of solution, while molality measures moles of solute per kilogram of solvent. Molarity is affected by temperature changes because liquid volume expands, whereas molality remains constant regardless of temperature.

Does adding more solvent change the number of moles of solute?

No, adding solvent only changes the concentration and the total volume of the solution. The total number of moles of solute remains constant during a dilution process, which is the basis for the M₁V₁ = M₂V₂ equation.

Why is molarity used more often than mass percent?

Molarity is preferred in laboratory settings because it relates directly to the stoichiometry of chemical reactions. Since chemists measure liquids by volume using graduated cylinders or pipettes, knowing the moles per liter allows for quick conversion to reaction ratios.

How do you convert grams to molarity?

To convert grams to molarity, first divide the mass of the solute by its molar mass to find the number of moles. Then, divide those moles by the total volume of the solution in liters.

Can molarity be greater than 1?

Yes, molarity can certainly be greater than 1; for example, concentrated laboratory acids like Sulfuric Acid can have a molarity as high as 18 M. The value depends entirely on the solubility of the substance and the amount dissolved.

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