MOLARITY WORKSHEET #1 | Exams Chemistry | Docsity - Free Printable

Here are the step-by-step solutions for each problem on the worksheet.

Problem 1

Question: If 10.7 grams of NH₄Cl is dissolved in enough water to make 800 mL of solution, what will be its molarity?

Step 1: Find the Molar Mass of NH₄Cl.
To do this, add up the atomic masses from the periodic table:
* Nitrogen (N): 14.01 g/mol
* Hydrogen (H): 1.01 g/mol × 4 atoms = 4.04 g/mol
* Chlorine (Cl): 35.45 g/mol
* Total Molar Mass = 14.01 + 4.04 + 35.45 = 53.50 g/mol

Step 2: Convert grams to moles.
Use the formula: $\text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}}$
$$ \text{Moles} = \frac{10.7 \text{ g}}{53.50 \text{ g/mol}} = 0.20 \text{ moles} $$

Step 3: Convert volume to Liters.
The volume is given in milliliters (mL), so divide by 1000.
$$ 800 \text{ mL} = 0.800 \text{ L} $$

Step 4: Calculate Molarity.
Use the formula: $\text{Molarity (M)} = \frac{\text{Moles}}{\text{Liters}}$
$$ M = \frac{0.20 \text{ mol}}{0.800 \text{ L}} = 0.25 \text{ mol/L} $$

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Problem 2

Question: Calculate the molarity of a solution prepared by dissolving 6.80 grams of AgNO₃ in enough water to make 2.50 liters of solution.

Step 1: Find the Molar Mass of AgNO₃.
* Silver (Ag): 107.87 g/mol
* Nitrogen (N): 14.01 g/mol
* Oxygen (O): 16.00 g/mol × 3 atoms = 48.00 g/mol
* Total Molar Mass = 107.87 + 14.01 + 48.00 = 169.88 g/mol

Step 2: Convert grams to moles.
$$ \text{Moles} = \frac{6.80 \text{ g}}{169.88 \text{ g/mol}} \approx 0.0400 \text{ moles} $$

Step 3: Calculate Molarity.
The volume is already in Liters (2.50 L).
$$ M = \frac{0.0400 \text{ mol}}{2.50 \text{ L}} = 0.016 \text{ mol/L} $$

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Problem 3

Question: How many moles of CaCl₂ are required to prepare 2.00 liters of 0.700 M CaCl₂?

Step 1: Identify the knowns.
* Molarity (M) = 0.700 mol/L
* Volume (V) = 2.00 L

Step 2: Rearrange the Molarity formula.
Since $M = \frac{\text{moles}}{\text{Liters}}$, we can multiply both sides by Liters to find moles:
$$ \text{Moles} = \text{Molarity} \times \text{Volume} $$

Step 3: Calculate.
$$ \text{Moles} = 0.700 \text{ mol/L} \times 2.00 \text{ L} = 1.4 \text{ moles} $$

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Problem 4

Question: What mass, in grams, of CaCl₂ will be required to prepare the above solution?

Step 1: Find the Molar Mass of CaCl₂.
* Calcium (Ca): 40.08 g/mol
* Chlorine (Cl): 35.45 g/mol × 2 atoms = 70.90 g/mol
* Total Molar Mass = 40.08 + 70.90 = 110.98 g/mol

Step 2: Convert moles to grams.
From Problem 3, we know we need 1.4 moles.
$$ \text{Mass} = \text{Moles} \times \text{Molar Mass} $$
$$ \text{Mass} = 1.4 \text{ mol} \times 110.98 \text{ g/mol} = 155.372 \text{ g} $$

Rounding to 3 significant figures gives 155 grams.

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Problem 5

Question: How many grams of KNO₃ will be required to prepare 800 mL of 1.40 M KNO₃?

Step 1: Convert volume to Liters.
$$ 800 \text{ mL} = 0.800 \text{ L} $$

Step 2: Calculate the moles needed.
$$ \text{Moles} = \text{Molarity} \times \text{Volume} $$
$$ \text{Moles} = 1.40 \text{ mol/L} \times 0.800 \text{ L} = 1.12 \text{ moles} $$

Step 3: Find the Molar Mass of KNO₃.
* Potassium (K): 39.10 g/mol
* Nitrogen (N): 14.01 g/mol
* Oxygen (O): 16.00 g/mol × 3 atoms = 48.00 g/mol
* Total Molar Mass = 39.10 + 14.01 + 48.00 = 101.11 g/mol

Step 4: Convert moles to grams.
$$ \text{Mass} = 1.12 \text{ mol} \times 101.11 \text{ g/mol} = 113.24 \text{ g} $$

Rounding to 3 significant figures gives 113 grams.

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Final Answer:
1. 0.25 mol/L
2. 0.016 mol/L
3. 1.4 moles
4. 155 grams
5. 113 grams