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Understanding Molarity
Before diving into practice problems, it’s essential to grasp the concept of molarity and its significance in chemistry.
What is Molarity?
Molarity (M) is defined as:
\[ \text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} \]
- Moles of solute: The amount of substance, measured in moles.
- Liters of solution: Total volume of the solution after solute is dissolved.
Why is Molarity Important?
- It allows chemists to prepare solutions with precise concentrations.
- Essential for stoichiometric calculations in reactions.
- Used in titrations, dilution calculations, and concentration adjustments.
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Basic Molarity Practice Problems
Let’s start with straightforward problems to build foundational understanding.
Problem 1: Calculating Molarity from Moles and Volume
Question:
What is the molarity of a solution prepared by dissolving 0.5 moles of sodium chloride (NaCl) in 2 liters of water?
Solution:
Using the formula:
\[ \text{M} = \frac{\text{moles}}{\text{liters}} \]
\[ \text{M} = \frac{0.5\, \text{mol}}{2\, \text{L}} = 0.25\, \text{M} \]
Answer:
The molarity of the NaCl solution is 0.25 M.
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Problem 2: Finding Moles of Solute from Molarity and Volume
Question:
A 3.0 M solution of sulfuric acid (H₂SO₄) has a volume of 500 mL. How many moles of H₂SO₄ are in the solution?
Solution:
Convert volume to liters:
\[ 500\, \text{mL} = 0.5\, \text{L} \]
Use the molarity formula:
\[ \text{moles} = \text{M} \times \text{volume} \]
\[ \text{moles} = 3.0\, \text{M} \times 0.5\, \text{L} = 1.5\, \text{mol} \]
Answer:
There are 1.5 moles of H₂SO₄ in the solution.
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Intermediate Molarity Practice Problems
As you progress, problems become more complex, involving dilution, solution preparation, and multiple steps.
Problem 3: Dilution Calculation
Question:
How much water must be added to 250 mL of a 2.0 M NaOH solution to dilute it to a 0.5 M solution?
Solution:
Use the dilution formula:
\[ C_1 V_1 = C_2 V_2 \]
Where:
- \( C_1 = 2.0\, \text{M} \)
- \( V_1 = 250\, \text{mL} \)
- \( C_2 = 0.5\, \text{M} \)
- \( V_2 = ? \)
Calculate \( V_2 \):
\[ V_2 = \frac{C_1 V_1}{C_2} = \frac{2.0\, \text{M} \times 250\, \text{mL}}{0.5\, \text{M}} = \frac{500}{0.5} = 1000\, \text{mL} \]
Determine the amount of water to add:
\[ \text{Water to add} = V_2 - V_1 = 1000\, \text{mL} - 250\, \text{mL} = 750\, \text{mL} \]
Answer:
Add 750 mL of water to dilute the solution to 0.5 M.
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Problem 4: Preparing a Solution of Known Molarity
Question:
How many grams of potassium permanganate (KMnO₄) are needed to prepare 250 mL of a 0.1 M solution?
Solution:
1. Calculate moles required:
\[ \text{moles} = \text{M} \times \text{volume} \]
\[ \text{volume} = 0.25\, \text{L} \]
\[ \text{moles} = 0.1\, \text{M} \times 0.25\, \text{L} = 0.025\, \text{mol} \]
2. Find molar mass of KMnO₄:
\[ \text{Molar mass} = 39.10\, (\text{K}) + 54.94\, (\text{Mn}) + 4 \times 16.00\, (\text{O}) = 158.04\, \text{g/mol} \]
3. Calculate grams:
\[ \text{grams} = \text{moles} \times \text{molar mass} \]
\[ \text{grams} = 0.025\, \text{mol} \times 158.04\, \text{g/mol} \approx 3.951\, \text{g} \]
Answer:
Approximately 3.95 grams of KMnO₄ are needed.
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Advanced Molarity Practice Problems
Now, let’s tackle problems that integrate multiple concepts, including titrations, molarity in reactions, and real-world applications.
Problem 5: Titration Calculation
Question:
In a titration, 25.0 mL of hydrochloric acid (HCl) of unknown concentration is neutralized by 30.0 mL of 0.10 M sodium hydroxide (NaOH). What is the molarity of the HCl solution?
Solution:
1. Write the balanced equation:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
2. Calculate moles of NaOH:
\[ \text{moles} = \text{M} \times \text{volume} \]
\[ = 0.10\, \text{M} \times 0.030\, \text{L} = 0.003\, \text{mol} \]
3. Moles of HCl are equal to moles of NaOH (1:1 ratio):
\[ \text{moles of HCl} = 0.003\, \text{mol} \]
4. Find molarity of HCl:
\[ \text{M} = \frac{\text{moles}}{\text{volume in liters}} = \frac{0.003\, \text{mol}}{0.025\, \text{L}} = 0.12\, \text{M} \]
Answer:
The molarity of the HCl solution is 0.12 M.
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Problem 6: Solution Preparation from Stock Solution
Question:
How would you prepare 1.0 L of a 0.2 M sodium carbonate (Na₂CO₃) solution from a 1.0 M stock solution?
Solution:
Use the dilution formula:
\[ C_1 V_1 = C_2 V_2 \]
Where:
- \( C_1 = 1.0\, \text{M} \)
- \( C_2 = 0.2\, \text{M} \)
- \( V_2 = 1.0\, \text{L} \)
- \( V_1 = ? \)
Calculate \( V_1 \):
\[ V_1 = \frac{C_2 V_2}{C_1} = \frac{0.2\, \text{M} \times 1.0\, \text{L}}{1.0\, \text{M}} = 0.2\, \text{L} \]
Convert to mL:
\[ 0.2\, \text{L} = 200\, \text{mL} \]
Answer:
Take 200 mL of the 1.0 M stock solution and dilute it to a total volume of 1.0 L with distilled water.
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Tips for Solving Molarity Problems Effectively
- Always convert units carefully: Ensure volume is in liters when using molarity.
- Understand the relationship: Moles, volume, and molarity are interconnected; use the formula \( \text{
Frequently Asked Questions
What is the definition of molarity in chemistry?
Molarity is a measure of concentration defined as the number of moles of solute dissolved in one liter of solution (mol/L).
How do you calculate the molarity of a solution given the mass of solute and volume of solution?
First, convert the mass of solute to moles using its molar mass, then divide that number by the volume of the solution in liters: Molarity = moles of solute / liters of solution.
What is the molarity of a solution made by dissolving 5 grams of NaCl in 2 liters of water?
First, find moles of NaCl: 5 g / 58.44 g/mol ≈ 0.0855 mol. Then, divide by 2 liters: 0.0855 mol / 2 L ≈ 0.0428 M. So, the molarity is approximately 0.043 M.
How do you prepare a 0.5 M NaOH solution if you have a stock solution of 2 M NaOH?
Use the dilution formula: C1V1 = C2V2. Rearranged as V1 = (C2 × V2) / C1. For example, to prepare 1 liter of 0.5 M solution: V1 = (0.5 M × 1 L) / 2 M = 0.25 L or 250 mL of the 2 M stock solution, diluted to 1 liter with water.
What is the molarity of a solution if 10 mL of it contains 0.02 mol of solute?
Convert 10 mL to liters: 0.010 L. Then, molarity = 0.02 mol / 0.010 L = 2 M.
If you dilute 100 mL of a 3 M solution to a total volume of 500 mL, what is the new molarity?
Use dilution formula: C1V1 = C2V2. C2 = (C1 × V1) / V2 = (3 M × 0.1 L) / 0.5 L = 0.6 M.
How do you find the molarity of a solution if you know the number of moles and volume in milliliters?
Convert volume to liters, then divide moles by liters: Molarity = moles / (volume in mL / 1000).
What are common mistakes to avoid when solving molarity practice problems?
Common mistakes include forgetting to convert units properly, mixing up moles and grams, not converting volume to liters, and neglecting to account for dilution factors.
Why is molarity preferred over other concentration units in solution chemistry problems?
Molarity is convenient because it relates directly to the amount of solute in a given volume, making calculations involving reactions, dilutions, and titrations straightforward and consistent.
