Chemistry worksheet focusing on molarity and molality calculations with various solute and solution problems.

Problem Set: Molarity and Molality Problems

Below, I will solve each problem step-by-step with clear explanations.

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#### Problem 1:
What is the molarity of an aqueous solution that contains 14.2 g NaCl dissolved in 2.35 L of solution?

Solution:
1. Molarity Formula:
\[
\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}}
\]

2. Calculate moles of NaCl:
The molar mass of NaCl is:
\[
\text{Molar mass of NaCl} = 22.99 \, \text{g/mol (Na)} + 35.45 \, \text{g/mol (Cl)} = 58.44 \, \text{g/mol}
\]
Using the given mass of NaCl (14.2 g):
\[
\text{Moles of NaCl} = \frac{\text{mass of NaCl}}{\text{molar mass of NaCl}} = \frac{14.2 \, \text{g}}{58.44 \, \text{g/mol}} \approx 0.243 \, \text{mol}
\]

3. Calculate molarity:
The volume of the solution is 2.35 L. Using the molarity formula:
\[
\text{Molarity} = \frac{\text{moles of NaCl}}{\text{volume of solution}} = \frac{0.243 \, \text{mol}}{2.35 \, \text{L}} \approx 0.103 \, \text{M}
\]

Answer:
\[
\boxed{0.103 \, \text{M}}
\]

---

#### Problem 2:
How many grams of NaCl would be contained in 1.00 L of the solution in #1?

Solution:
1. From Problem 1:
The molarity of the solution is \(0.103 \, \text{M}\). This means there are 0.103 moles of NaCl per liter of solution.

2. Calculate moles of NaCl in 1.00 L:
Since the molarity is \(0.103 \, \text{M}\), in 1.00 L of solution:
\[
\text{Moles of NaCl} = 0.103 \, \text{mol}
\]

3. Convert moles to grams:
Using the molar mass of NaCl (\(58.44 \, \text{g/mol}\)):
\[
\text{Mass of NaCl} = \text{moles of NaCl} \times \text{molar mass of NaCl} = 0.103 \, \text{mol} \times 58.44 \, \text{g/mol} \approx 6.02 \, \text{g}
\]

Answer:
\[
\boxed{6.02 \, \text{g}}
\]

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#### Problem 3:
A solution is made by dissolving 17.0 g of lithium iodide (LiI) in enough water to make 387 mL of solution. What is the molarity of the solution?

Solution:
1. Molarity Formula:
\[
\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}}
\]

2. Calculate moles of LiI:
The molar mass of LiI is:
\[
\text{Molar mass of LiI} = 6.94 \, \text{g/mol (Li)} + 126.90 \, \text{g/mol (I)} = 133.84 \, \text{g/mol}
\]
Using the given mass of LiI (17.0 g):
\[
\text{Moles of LiI} = \frac{\text{mass of LiI}}{\text{molar mass of LiI}} = \frac{17.0 \, \text{g}}{133.84 \, \text{g/mol}} \approx 0.127 \, \text{mol}
\]

3. Convert volume to liters:
The volume of the solution is 387 mL. Convert to liters:
\[
387 \, \text{mL} = 0.387 \, \text{L}
\]

4. Calculate molarity:
Using the molarity formula:
\[
\text{Molarity} = \frac{\text{moles of LiI}}{\text{volume of solution}} = \frac{0.127 \, \text{mol}}{0.387 \, \text{L}} \approx 0.328 \, \text{M}
\]

Answer:
\[
\boxed{0.328 \, \text{M}}
\]

---

#### Problem 4:
How many grams are contained in 150 mL of a 0.20 M solution of K₂SO₄?

Solution:
1. Molarity Formula:
\[
\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}}
\]

2. Convert volume to liters:
The volume of the solution is 150 mL. Convert to liters:
\[
150 \, \text{mL} = 0.150 \, \text{L}
\]

3. Calculate moles of K₂SO₄:
The molarity of the solution is \(0.20 \, \text{M}\). Using the molarity formula:
\[
\text{Moles of K₂SO₄} = \text{Molarity} \times \text{volume of solution} = 0.20 \, \text{M} \times 0.150 \, \text{L} = 0.030 \, \text{mol}
\]

4. Calculate mass of K₂SO₄:
The molar mass of K₂SO₄ is:
\[
\text{Molar mass of K₂SO₄} = 2 \times 39.10 \, \text{g/mol (K)} + 32.07 \, \text{g/mol (S)} + 4 \times 16.00 \, \text{g/mol (O)} = 174.27 \, \text{g/mol}
\]
Using the moles of K₂SO₄:
\[
\text{Mass of K₂SO₄} = \text{moles of K₂SO₄} \times \text{molar mass of K₂SO₄} = 0.030 \, \text{mol} \times 174.27 \, \text{g/mol} \approx 5.23 \, \text{g}
\]

Answer:
\[
\boxed{5.23 \, \text{g}}
\]

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#### Problem 5:
Calculate the molarity of a water solution of CaCl₂, given that 5.04 L of the solution contains 612 g of CaCl₂.

Solution:
1. Molarity Formula:
\[
\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}}
\]

2. Calculate moles of CaCl₂:
The molar mass of CaCl₂ is:
\[
\text{Molar mass of CaCl₂} = 40.08 \, \text{g/mol (Ca)} + 2 \times 35.45 \, \text{g/mol (Cl)} = 110.98 \, \text{g/mol}
\]
Using the given mass of CaCl₂ (612 g):
\[
\text{Moles of CaCl₂} = \frac{\text{mass of CaCl₂}}{\text{molar mass of CaCl₂}} = \frac{612 \, \text{g}}{110.98 \, \text{g/mol}} \approx 5.52 \, \text{mol}
\]

3. Calculate molarity:
The volume of the solution is 5.04 L. Using the molarity formula:
\[
\text{Molarity} = \frac{\text{moles of CaCl₂}}{\text{volume of solution}} = \frac{5.52 \, \text{mol}}{5.04 \, \text{L}} \approx 1.10 \, \text{M}
\]

Answer:
\[
\boxed{1.10 \, \text{M}}
\]

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#### Problem 6:
Suppose you wish to make 0.85 L of a 0.25 M solution of silver nitrate, AgNO₃. How many grams of AgNO₃ would you need to prepare the solution?

Solution:
1. Molarity Formula:
\[
\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}}
\]

2. Calculate moles of AgNO₃:
The molarity of the solution is \(0.25 \, \text{M}\), and the volume is 0.85 L. Using the molarity formula:
\[
\text{Moles of AgNO₃} = \text{Molarity} \times \text{volume of solution} = 0.25 \, \text{M} \times 0.85 \, \text{L} = 0.2125 \, \text{mol}
\]

3. Calculate mass of AgNO₃:
The molar mass of AgNO₃ is:
\[
\text{Molar mass of AgNO₃} = 107.87 \, \text{g/mol (Ag)} + 14.01 \, \text{g/mol (N)} + 3 \times 16.00 \, \text{g/mol (O)} = 169.87 \, \text{g/mol}
\]
Using the moles of AgNO₃:
\[
\text{Mass of AgNO₃} = \text{moles of AgNO₃} \times \text{molar mass of AgNO₃} = 0.2125 \, \text{mol} \times 169.87 \, \text{g/mol} \approx 36.0 \, \text{g}
\]

Answer:
\[
\boxed{36.0 \, \text{g}}
\]

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Final Answers:

1. \(\boxed{0.103 \, \text{M}}\)
2. \(\boxed{6.02 \, \text{g}}\)
3. \(\boxed{0.328 \, \text{M}}\)
4. \(\boxed{5.23 \, \text{g}}\)
5. \(\boxed{1.10 \, \text{M}}\)
6. \(\boxed{36.0 \, \text{g}}\)