Hard Solution Preparation Practice Questions

Concept Explanation

Solution preparation is the precise laboratory process of creating a liquid mixture where a solute is dissolved in a solvent to achieve a specific concentration, often requiring complex calculations involving molarity, density, and mass percentages. Mastering this involves more than just simple mixing; it requires an understanding of what molarity is and how to manipulate units to achieve high precision. In advanced chemistry, preparing solutions often involves diluting concentrated stock acids, accounting for hydrate weights, or converting between molarity and molality. Precise preparation is critical in fields like analytical chemistry and pharmacology, where even a 1% error can invalidate experimental results or compromise safety. Key formulas include the dilution equation (C1V1 = C2V2) and mass-based calculations using the molecular formula of the solute to determine the exact mass needed.

Solved Examples

These examples demonstrate the multi-step reasoning required for hard solution preparation problems.

Example 1: Preparing a Solution from a Hydrate
Prepare 500.0 mL of a 0.250 M Copper(II) Sulfate solution using Copper(II) Sulfate Pentahydrate (CuSO₄·5H₂O). The molar mass of the hydrate is 249.68 g/mol.

1. Calculate the moles of CuSO₄ needed: Moles = Molarity × Volume (L) = 0.250 mol/L × 0.500 L = 0.125 moles.

2. Since 1 mole of the hydrate provides 1 mole of CuSO₄, we need 0.125 moles of CuSO₄·5H₂O.

3. Convert moles to grams: mass = 0.125 mol × 249.68 g/mol = 31.21 grams.

4. Dissolve 31.21 g of the hydrate in some distilled water, then dilute to the 500 mL mark in a volumetric flask.

Example 2: Stock Acid Dilution
Concentrated Sulfuric Acid (H₂SO₄) is 18.0 M. How would you prepare 2.0 L of a 0.50 M H₂SO₄ solution?

1. Use the dilution formula: C₁V₁ = C₂V₂.

2. (18.0 M)(V₁) = (0.50 M)(2.0 L).

3. V₁ = (0.50 × 2.0) / 18.0 = 0.0556 L or 55.6 mL.

4. Safety Step: Always add acid to water. Measure 55.6 mL of stock acid and slowly add it to approximately 1.5 L of water, then dilute to exactly 2.0 L.

Example 3: Density and Mass Percentage
A bottle of concentrated HCl is 37.0% by mass and has a density of 1.19 g/mL. Calculate the molarity of this stock solution.

1. Assume 1.00 L (1000 mL) of solution. Mass = density × volume = 1.19 g/mL × 1000 mL = 1190 g.

2. Calculate mass of HCl: 1190 g × 0.37 = 440.3 g HCl.

3. Convert mass to moles: 440.3 g / 36.46 g/mol = 12.08 moles.

4. Since this is in 1 L, the molarity is 12.1 M.

Practice Questions

Test your skills with these Hard Solution Preparation Practice Questions. Ensure you have a periodic table and calculator ready.

1. You need to prepare 250.0 mL of a 0.150 M solution of Silver Nitrate (AgNO₃). How many grams of AgNO₃ are required?

2. A chemist needs to prepare 1.50 L of 0.200 M KOH. The available KOH is 85% pure by mass. How many grams of this impure KOH are needed?

3. Calculate the final concentration when 150.0 mL of 0.500 M NaCl is mixed with 350.0 mL of 0.200 M NaCl.

4. How would you prepare 100.0 mL of 0.050 M HCl starting from a 12.0 M stock solution? Specify the volume of stock and water needed.

5. What mass of Na₂SO₄·10H₂O (molar mass = 322.2 g/mol) is needed to prepare 750 mL of a solution that is 0.100 M in Na⁺ ions?

6. A solution is prepared by dissolving 25.0 g of glucose (C₆H₁₂O₆) in 200.0 g of water. The final solution density is 1.04 g/mL. Calculate the molarity.

7. You have a 5.00 M stock solution of NaOH. You dilute 10.0 mL of this to 250.0 mL (Solution A). Then you take 50.0 mL of Solution A and dilute it to 500.0 mL (Solution B). What is the concentration of Solution B?

8. Concentrated Nitric Acid (HNO₃) is 70.0% by mass with a density of 1.42 g/mL. How many milliliters of this acid are required to prepare 500.0 mL of 2.00 M HNO₃?

9. How much water must be added to 25.0 mL of 6.00 M HCl to make it 0.500 M?

10. A 2.50 M solution of MgCl₂ has a density of 1.15 g/mL. What is the molality of this solution?

Answers & Explanations

Detailed solutions for the Hard Solution Preparation Practice Questions are provided below.

1. 6.37 g: Moles = 0.250 L × 0.150 mol/L = 0.0375 mol. Mass = 0.0375 mol × 169.87 g/mol (AgNO₃) = 6.37 g.

2. 19.8 g: Moles needed = 1.50 L × 0.200 M = 0.300 mol. Mass of pure KOH = 0.300 mol × 56.11 g/mol = 16.833 g. Since it is 85% pure: 16.833 / 0.85 = 19.8 g.

3. 0.290 M: Total moles = (0.150 L × 0.500 M) + (0.350 L × 0.200 M) = 0.075 + 0.070 = 0.145 mol. Total volume = 0.500 L. Molarity = 0.145 / 0.500 = 0.290 M.

4. 0.42 mL stock, 99.58 mL water: V₁ = (0.050 M × 100 mL) / 12.0 M = 0.4167 mL. Add 0.42 mL stock to 99.58 mL water.

5. 12.1 g: We need 0.100 M Na⁺. Since 1 mol Na₂SO₄ gives 2 mol Na⁺, the concentration of Na₂SO₄ must be 0.050 M. Moles = 0.750 L × 0.050 M = 0.0375 mol. Mass = 0.0375 mol × 322.2 g/mol = 12.1 g.

6. 0.641 M: Total mass = 25.0 + 200.0 = 225.0 g. Volume = mass / density = 225.0 / 1.04 = 216.35 mL (0.21635 L). Moles glucose = 25.0 / 180.16 = 0.1387 mol. Molarity = 0.1387 / 0.21635 = 0.641 M.

7. 0.020 M: Sol A = (5.00 M × 10 mL) / 250 mL = 0.200 M. Sol B = (0.200 M × 50 mL) / 500 mL = 0.020 M.

8. 63.4 mL: Stock Molarity = (1.42 g/mL × 1000 mL × 0.70) / 63.01 g/mol = 15.77 M. V₁ = (2.00 M × 500 mL) / 15.77 M = 63.4 mL.

9. 275 mL: V₂ = (6.00 M × 25.0 mL) / 0.500 M = 300 mL. Water added = 300 - 25 = 275 mL.

10. 2.74 m: 1 L solution = 1150 g. Mass MgCl₂ = 2.50 mol × 95.21 g/mol = 238.03 g. Mass water = 1150 - 238.03 = 911.97 g (0.912 kg). Molality = 2.50 mol / 0.912 kg = 2.74 m.

Quick Quiz

Frequently Asked Questions

What is the difference between molarity and molality in solution preparation?

Molarity measures moles of solute per liter of solution and is temperature-dependent because volume changes with heat. Molality measures moles of solute per kilogram of solvent, making it independent of temperature and ideal for boiling point or freezing point calculations.

Why must acid be added to water instead of water to acid?

Adding water to a concentrated acid can cause a violent exothermic reaction that splashes corrosive liquid due to the heat generated at the interface. Adding acid to water allows the large volume of water to absorb the heat safely, preventing boiling and splattering.

How do I account for the purity of a chemical when preparing a solution?

To account for purity, divide the calculated theoretical mass by the decimal form of the purity percentage. For example, if you need 10 grams of a substance that is only 90% pure, you must weigh out 11.11 grams (10 / 0.90) to ensure you have 10 grams of the actual chemical.

Does the volume of a solute affect the final volume of a solution?

Yes, the solute occupies space, which is why you should never add the full volume of solvent to the solute. Instead, dissolve the solute in a smaller portion of solvent first and then "top off" to the final volume mark in a volumetric flask.

What is a stock solution in laboratory chemistry?

A stock solution is a highly concentrated, stable solution used to prepare more dilute working solutions through the process of dilution. This method saves storage space and improves accuracy compared to weighing out tiny amounts of solid for every individual trial.

How do you convert mass percentage to molarity?

To convert mass percent to molarity, multiply the density (g/mL) by 1000 to get the mass of 1 L of solution, then multiply by the mass percentage (as a decimal) to find the mass of the solute. Finally, divide that mass by the solute's molar mass to find the moles per liter.

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