Chemistry worksheet on concentrations and dilutions, featuring molarity calculations and dilution problems.

Problem: Concentrations and Dilutions Worksheet

Below, I will solve each problem step by step, explaining the reasoning and formulas used.

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#### 1. A solution is made by adding 27.5 g of calcium fluoride (CaF₂) to enough water to make 1.00 L. What is the concentration (molarity)?

Solution:
To find the molarity (\( M \)) of the solution, we use the formula:
\[
M = \frac{\text{moles of solute}}{\text{liters of solution}}
\]

Step 1: Calculate the moles of CaF₂.
The molar mass of CaF₂ is calculated as follows:
- Calcium (Ca): \( 40.08 \, \text{g/mol} \)
- Fluorine (F): \( 19.00 \, \text{g/mol} \times 2 = 38.00 \, \text{g/mol} \)
- Total molar mass of CaF₂: \( 40.08 + 38.00 = 78.08 \, \text{g/mol} \)

Now, convert grams of CaF₂ to moles:
\[
\text{Moles of CaF₂} = \frac{\text{mass of CaF₂}}{\text{molar mass of CaF₂}} = \frac{27.5 \, \text{g}}{78.08 \, \text{g/mol}} \approx 0.352 \, \text{mol}
\]

Step 2: Calculate the molarity.
The volume of the solution is given as 1.00 L. Using the molarity formula:
\[
M = \frac{\text{moles of solute}}{\text{liters of solution}} = \frac{0.352 \, \text{mol}}{1.00 \, \text{L}} = 0.352 \, \text{M}
\]

Answer:
\[
\boxed{0.352 \, \text{M}}
\]

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#### 2. What is the concentration of each ion in solution #1? (Hint: first write an equation)

Solution:
From problem #1, the molarity of CaF₂ is 0.352 M. Calcium fluoride dissociates in water as follows:
\[
\text{CaF}_2 \rightarrow \text{Ca}^{2+} + 2\text{F}^-
\]

From the balanced equation, we see that:
- 1 mole of CaF₂ produces 1 mole of \(\text{Ca}^{2+}\) ions.
- 1 mole of CaF₂ produces 2 moles of \(\text{F}^-\) ions.

Thus:
- The concentration of \(\text{Ca}^{2+}\) ions is the same as the molarity of CaF₂:
\[
[\text{Ca}^{2+}] = 0.352 \, \text{M}
\]
- The concentration of \(\text{F}^-\) ions is twice the molarity of CaF₂:
\[
[\text{F}^-] = 2 \times 0.352 \, \text{M} = 0.704 \, \text{M}
\]

Answer:
\[
\boxed{[\text{Ca}^{2+}] = 0.352 \, \text{M}, \, [\text{F}^-] = 0.704 \, \text{M}}
\]

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#### 3. How much solute is contained in 500 mL of a 2.5 M solution? (Hint: Liters)

Solution:
We use the formula for molarity:
\[
M = \frac{\text{moles of solute}}{\text{liters of solution}}
\]
Rearranging to solve for moles of solute:
\[
\text{moles of solute} = M \times \text{liters of solution}
\]

Convert 500 mL to liters:
\[
500 \, \text{mL} = 0.500 \, \text{L}
\]

Now calculate the moles of solute:
\[
\text{moles of solute} = 2.5 \, \text{M} \times 0.500 \, \text{L} = 1.25 \, \text{mol}
\]

Answer:
\[
\boxed{1.25 \, \text{mol}}
\]

---

#### 4. A student needs 250 mL of a 0.75 M solution of sodium acetate. How many moles of sodium acetate are needed?

Solution:
Again, we use the formula for molarity:
\[
M = \frac{\text{moles of solute}}{\text{liters of solution}}
\]
Rearranging to solve for moles of solute:
\[
\text{moles of solute} = M \times \text{liters of solution}
\]

Convert 250 mL to liters:
\[
250 \, \text{mL} = 0.250 \, \text{L}
\]

Now calculate the moles of sodium acetate:
\[
\text{moles of sodium acetate} = 0.75 \, \text{M} \times 0.250 \, \text{L} = 0.1875 \, \text{mol}
\]

Answer:
\[
\boxed{0.1875 \, \text{mol}}
\]

---

#### 5. A student has 12.0 M HCl and needs to make 1.0 L of 2.0 M HCl. What volume of the concentrated acid is needed?

Solution:
We use the dilution formula:
\[
M_1 V_1 = M_2 V_2
\]
where:
- \( M_1 \) = initial molarity (12.0 M)
- \( V_1 \) = initial volume (unknown)
- \( M_2 \) = final molarity (2.0 M)
- \( V_2 \) = final volume (1.0 L)

Rearranging to solve for \( V_1 \):
\[
V_1 = \frac{M_2 V_2}{M_1}
\]

Substitute the values:
\[
V_1 = \frac{2.0 \, \text{M} \times 1.0 \, \text{L}}{12.0 \, \text{M}} = \frac{2.0}{12.0} = 0.1667 \, \text{L}
\]

Convert to milliliters:
\[
0.1667 \, \text{L} = 166.7 \, \text{mL}
\]

Answer:
\[
\boxed{166.7 \, \text{mL}}
\]

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#### 6. Another student adds 500 mL of H₂O to the solution made in #5. What is the concentration of the new solution? (Assume the volumes are additive.)

Solution:
From problem #5, the student made 1.0 L of 2.0 M HCl. Now, 500 mL of water is added, so the total volume becomes:
\[
\text{Total volume} = 1.0 \, \text{L} + 0.500 \, \text{L} = 1.5 \, \text{L}
\]

The number of moles of HCl remains the same (since no solute is added or removed). From problem #5, the moles of HCl in 1.0 L of 2.0 M solution is:
\[
\text{moles of HCl} = 2.0 \, \text{M} \times 1.0 \, \text{L} = 2.0 \, \text{mol}
\]

Now, calculate the new molarity:
\[
M = \frac{\text{moles of solute}}{\text{liters of solution}} = \frac{2.0 \, \text{mol}}{1.5 \, \text{L}} \approx 1.33 \, \text{M}
\]

Answer:
\[
\boxed{1.33 \, \text{M}}
\]

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#### 7. A third student wants to dilute the solution in #6 to 1.0 M. How much water must be added? (Assume the volumes are additive.)

Solution:
We use the dilution formula again:
\[
M_1 V_1 = M_2 V_2
\]
where:
- \( M_1 \) = initial molarity (1.33 M)
- \( V_1 \) = initial volume (1.5 L)
- \( M_2 \) = final molarity (1.0 M)
- \( V_2 \) = final volume (unknown)

Rearranging to solve for \( V_2 \):
\[
V_2 = \frac{M_1 V_1}{M_2}
\]

Substitute the values:
\[
V_2 = \frac{1.33 \, \text{M} \times 1.5 \, \text{L}}{1.0 \, \text{M}} = 1.995 \, \text{L}
\]

The volume of water to be added is:
\[
\text{Volume of water added} = V_2 - V_1 = 1.995 \, \text{L} - 1.5 \, \text{L} = 0.495 \, \text{L}
\]

Convert to milliliters:
\[
0.495 \, \text{L} = 495 \, \text{mL}
\]

Answer:
\[
\boxed{495 \, \text{mL}}
\]

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#### 8. If 95.0 mL of rubbing alcohol are added to enough water to make 150.0 mL, what is the volume %?

Solution:
The volume percentage (\% v/v) is calculated using the formula:
\[
\text{Volume \%} = \left( \frac{\text{volume of solute}}{\text{total volume of solution}} \right) \times 100
\]

Here:
- Volume of solute (rubbing alcohol) = 95.0 mL
- Total volume of solution = 150.0 mL

Substitute the values:
\[
\text{Volume \%} = \left( \frac{95.0 \, \text{mL}}{150.0 \, \text{mL}} \right) \times 100 \approx 63.33\%
\]

Answer:
\[
\boxed{63.33\%}
\]

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#### 9. Determine the mass % of a NaCl solution if 58.5 grams of NaCl was dissolved in 50 mL of water (assume the density of water to be 1 g/mL).

Solution:
The mass percentage (\% m/m) is calculated using the formula:
\[
\text{Mass \%} = \left( \frac{\text{mass of solute}}{\text{total mass of solution}} \right) \times 100
\]

First, calculate the mass of water:
\[
\text{Mass of water} = \text{volume of water} \times \text{density of water} = 50 \, \text{mL} \times 1 \, \text{g/mL} = 50 \, \text{g}
\]

Next, calculate the total mass of the solution:
\[
\text{Total mass of solution} = \text{mass of solute} + \text{mass of solvent} = 58.5 \, \text{g} + 50 \, \text{g} = 108.5 \, \text{g}
\]

Now, calculate the mass percentage:
\[
\text{Mass \%} = \left( \frac{58.5 \, \text{g}}{108.5 \, \text{g}} \right) \times 100 \approx 53.92\%
\]

Answer:
\[
\boxed{53.92\%}
\]

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#### 10. If we wished to prepare 400 mL of a 10% by mass NaCl solution, what mass of NaCl would we use? Density of the solution is 1.05 g/mL.

Solution:
For a 10% by mass solution, the mass of solute (NaCl) is 10% of the total mass of the solution. First, calculate the total mass of the solution using its density:
\[
\text{Total mass of solution} = \text{volume of solution} \times \text{density of solution} = 400 \, \text{mL} \times 1.05 \, \text{g/mL} = 420 \, \text{g}
\]

Now, calculate the mass of NaCl:
\[
\text{Mass of NaCl} = \text{mass \%} \times \text{total mass of solution} = 0.10 \times 420 \, \text{g} = 42 \, \text{g}
\]

Answer:
\[
\boxed{42 \, \text{g}}
\]

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Final Answers:

1. \(\boxed{0.352 \, \text{M}}\)
2. \(\boxed{[\text{Ca}^{2+}] = 0.352 \, \text{M}, \, [\text{F}^-] = 0.704 \, \text{M}}\)
3. \(\boxed{1.25 \, \text{mol}}\)
4. \(\boxed{0.1875 \, \text{mol}}\)
5. \(\boxed{166.7 \, \text{mL}}\)
6. \(\boxed{1.33 \, \text{M}}\)
7. \(\boxed{495 \, \text{mL}}\)
8. \(\boxed{63.33\%}\)
9. \(\boxed{53.92\%}\)
10. \(\boxed{42 \, \text{g}}\)