Hard Dilution Practice Questions

Concept Explanation

Dilution is the process of decreasing the concentration of a solute in a solution by adding more solvent, usually without adding more solute. This fundamental chemical procedure relies on the principle of conservation of mass, where the total number of moles of solute remains constant even as the volume of the liquid increases. In laboratory settings, scientists use the dilution equation, M₁V₁ = M₂V₂, to calculate the relationships between initial and final molarities and volumes. Mastering hard dilution practice questions requires a deep understanding of how to manipulate these variables, especially when dealing with density, mass percentages, or multi-step serial dilutions. For those looking to strengthen their foundation, reviewing what is molarity can provide essential context before tackling these complex scenarios.

When solving advanced problems, it is critical to ensure that units are consistent. While the dilution formula works with any volume unit (milliliters, liters, or even microliters), both sides of the equation must match. In complex industrial or medical applications, dilutions are often expressed as ratios or parts-per-million (ppm). According to Khan Academy, molarity is the most common way to express concentration in these calculations. If you find yourself struggling with the underlying math, you might benefit from practicing grams to moles practice questions to ensure your unit conversions are flawless.

Solved Examples

The following examples demonstrate how to approach hard dilution practice questions involving density and multi-step processes.

1. Example 1: Concentration from Density
   A stock solution of sulfuric acid (H₂SO₄) has a density of 1.84 g/mL and is 98% H₂SO₄ by mass. What volume of this stock solution is required to prepare 5.0 L of a 0.50 M H₂SO₄ solution?
   

   1. Calculate the mass of 1 liter of stock: 1000 mL × 1.84 g/mL = 1840 g.

   2. Calculate the mass of pure H₂SO₄: 1840 g × 0.98 = 1803.2 g.

   3. Convert mass to moles (Molar mass of H₂SO₄ ≈ 98.08 g/mol): 1803.2 g / 98.08 g/mol = 18.38 moles. Thus, M₁ = 18.38 M.

   4. Use M₁V₁ = M₂V₂: (18.38 M)(V₁) = (0.50 M)(5.0 L).

   5. V₁ = 0.136 L or 136 mL.

2. Example 2: Serial Dilution
   A scientist takes 10.0 mL of a 2.0 M HCl solution and dilutes it to 100.0 mL (Solution A). Then, 5.0 mL of Solution A is diluted to 250.0 mL (Solution B). What is the final molarity of Solution B?
   

   1. Find Molarity of A: (2.0 M)(10.0 mL) = (M₂)(100.0 mL) → M₂ = 0.20 M.

   2. Find Molarity of B: (0.20 M)(5.0 mL) = (M₂)(250.0 mL).

   3. M₂ = 1.0 / 250 = 0.0040 M.

3. Example 3: Mixing and Diluting
   If 50.0 mL of 1.5 M NaOH is mixed with 30.0 mL of 0.5 M NaOH and the resulting mixture is diluted to a final volume of 500.0 mL, what is the final concentration?
   

   1. Calculate total moles: (0.050 L × 1.5 M) + (0.030 L × 0.5 M) = 0.075 + 0.015 = 0.090 moles.

   2. Apply final volume: M = 0.090 moles / 0.500 L.

   3. Final Molarity = 0.18 M.

Practice Questions

1. A concentrated nitric acid solution is 70.0% HNO₃ by mass and has a density of 1.42 g/mL. How many milliliters of this solution are needed to prepare 250 mL of 1.20 M HNO₃?

2. A 50.0 mL sample of 12.0 M HCl is diluted to 500.0 mL. Then, 20.0 mL of this new solution is further diluted to 1.00 L. Calculate the final molarity.

3. Calculate the volume of water that must be added to 150 mL of a 4.5 M KOH solution to result in a 0.75 M solution.

4. A glucose stock solution is 25.0% by mass with a density of 1.10 g/mL. If 15.0 mL of this stock is diluted to 100.0 mL, what is the final molarity? (Molar mass of glucose = 180.16 g/mol).

5. A 2.5 M stock solution is used to create a series of five 1:10 dilutions. What is the molarity of the final solution?

6. You have 500 mL of 0.20 M NaCl. You accidentally evaporate 150 mL of the water. What is the new concentration?

7. How much 18.0 M H₂SO₄ is required to make 2.0 L of a 0.10 M solution? Express your answer in microliters (µL).

8. A chemist mixes 100 mL of 0.5 M HCl with 200 mL of 0.1 M HCl and then dilutes the entire mixture to 1.0 L. What is the final concentration?

9. A solution is prepared by dissolving 25.0 g of BaCl₂ in enough water to make 250 mL. If 10.0 mL of this solution is diluted to 500 mL, what is the final molar concentration of Cl⁻ ions?

10. Find the final molarity when 15 mL of 6.0 M H₂SO₄ is added to 45 mL of 2.0 M H₂SO₄ and the total volume is adjusted to 150 mL.

Answers & Explanations

1. Answer: 19.0 mL
   First, find M₁: (1.42 g/mL × 1000 mL × 0.70) / 63.01 g/mol = 15.77 M. Using M₁V₁ = M₂V₂: (15.77 M)(V₁) = (1.20 M)(250 mL). V₁ = 19.02 mL.

2. Answer: 0.024 M
   First dilution: (12.0)(50) = (M₂)(500) → M₂ = 1.2 M. Second dilution: (1.2)(20) = (M₂)(1000) → M₂ = 0.024 M.

3. Answer: 750 mL
   M₁V₁ = M₂V₂: (4.5)(150) = (0.75)(V₂). V₂ = 900 mL. Since we started with 150 mL, the volume added is 900 - 150 = 750 mL.

4. Answer: 0.229 M
   M₁ = (1.10 g/mL × 1000 mL × 0.25) / 180.16 g/mol = 1.526 M. Dilution: (1.526)(15) = (M₂)(100). M₂ = 0.2289 M.

5. Answer: 2.5 × 10⁻⁵ M
   Each 1:10 dilution reduces concentration by a factor of 10. Five dilutions = 10⁵ reduction. 2.5 / 100,000 = 0.000025 M.

6. Answer: 0.286 M
   Moles = 0.500 L × 0.20 M = 0.10 moles. New volume = 500 - 150 = 350 mL (0.350 L). M = 0.10 / 0.350 = 0.286 M.

7. Answer: 11,111 µL
   (18.0)(V₁) = (0.10)(2.0 L). V₁ = 0.01111 L. 0.01111 L × 1,000,000 µL/L = 11,111 µL.

8. Answer: 0.07 M
   Moles = (0.1 × 0.5) + (0.2 × 0.1) = 0.05 + 0.02 = 0.07 moles. M = 0.07 moles / 1.0 L = 0.07 M.

9. Answer: 0.0192 M
   Moles BaCl₂ = 25.0 / 208.23 = 0.120 mol. M₁ = 0.120 / 0.250 = 0.480 M. After dilution: (0.48)(10) = (M₂)(500) → M₂ = 0.0096 M BaCl₂. Since there are 2 Cl⁻ per BaCl₂, [Cl⁻] = 0.0096 × 2 = 0.0192 M.

10. Answer: 1.2 M
    Total moles = (0.015 × 6.0) + (0.045 × 2.0) = 0.09 + 0.09 = 0.18 moles. M = 0.18 / 0.150 = 1.2 M.

Quick Quiz

Frequently Asked Questions

What is the difference between dilution and concentration?

Dilution involves adding more solvent to decrease the solute's concentration, while concentration involves removing solvent or adding more solute to increase the ratio of solute to solvent. Both processes alter the molarity but only adding/removing solute changes the total number of moles present.

Can I use milliliters instead of liters in M₁V₁ = M₂V₂?

Yes, you can use any volume unit as long as you are consistent on both sides of the equation. If V₁ is in mL, V₂ must also be in mL for the calculation to remain accurate.

Why do we add acid to water instead of water to acid during dilution?

Adding acid to water is a safety protocol because the hydration of strong acids is highly exothermic. Water has a high heat capacity and can absorb the heat generated, preventing the mixture from boiling and splashing acid onto the chemist, a common topic in molarity mistake guides.

What is a serial dilution?

A serial dilution is a stepwise dilution of a substance in solution where the dilution factor is kept constant at each step. This method is used to create very dilute solutions accurately, as seen in Wikipedia's overview of laboratory techniques.

How does temperature affect dilution calculations?

Temperature can affect the volume of a liquid due to thermal expansion or contraction, which slightly alters the molarity. For high-precision work, chemists use volumetric flasks calibrated at 20°C to minimize errors related to temperature fluctuations.

Does the mass of the solute change during dilution?

No, the mass of the solute remains constant because no solute is added or removed during the process. Only the volume of the solvent changes, which is why the moles of solute before and after dilution are equal.

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