Chemistry worksheet featuring concentration calculation problems with step-by-step solutions.

Problem Analysis:

The task involves solving concentration and volume problems related to solutions. The key formula used here is:

\[
\text{Concentration (c)} = \frac{\text{Number of moles (n)}}{\text{Volume (v)}}
\]

Where:
- \( c \) is the concentration in moles per cubic decimeter (\( \text{mol dm}^{-3} \)).
- \( n \) is the number of moles.
- \( v \) is the volume in cubic decimeters (\( \text{dm}^3 \)).

Additionally, we use the relationship:
\[
v = \frac{n}{c}
\]
to find the volume when the number of moles and concentration are known.

Let's solve each problem step by step.

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Problem 4:

What is the concentration of a solution containing 2.50 moles of sodium chloride in 500 cm³ of water?

#### Step 1: Convert the volume from cm³ to dm³.
\[
1 \, \text{dm}^3 = 1000 \, \text{cm}^3
\]
\[
500 \, \text{cm}^3 = \frac{500}{1000} \, \text{dm}^3 = 0.500 \, \text{dm}^3
\]

#### Step 2: Use the concentration formula.
\[
c = \frac{n}{v} = \frac{2.50 \, \text{moles}}{0.500 \, \text{dm}^3} = 5.00 \, \text{mol dm}^{-3}
\]

#### Final Answer:
\[
\boxed{5.00 \, \text{mol dm}^{-3}}
\]

---

Problem 5:

What is the concentration of a solution containing 0.300 moles of potassium hydroxide in 120 cm³ of water?

#### Step 1: Convert the volume from cm³ to dm³.
\[
120 \, \text{cm}^3 = \frac{120}{1000} \, \text{dm}^3 = 0.120 \, \text{dm}^3
\]

#### Step 2: Use the concentration formula.
\[
c = \frac{n}{v} = \frac{0.300 \, \text{moles}}{0.120 \, \text{dm}^3} = 2.50 \, \text{mol dm}^{-3}
\]

#### Final Answer:
\[
\boxed{2.50 \, \text{mol dm}^{-3}}
\]

---

Problem 6:

What is the concentration of a solution containing 10.0 moles of nitric acid in 20.0 dm³ of water?

#### Step 1: Use the concentration formula directly (no conversion needed since the volume is already in dm³).
\[
c = \frac{n}{v} = \frac{10.0 \, \text{moles}}{20.0 \, \text{dm}^3} = 0.500 \, \text{mol dm}^{-3}
\]

#### Final Answer:
\[
\boxed{0.500 \, \text{mol dm}^{-3}}
\]

---

Problem 7:

What volume is required to make a 2.00 mol dm⁻³ solution starting with 5.00 moles of lead nitrate?

#### Step 1: Use the formula for volume.
\[
v = \frac{n}{c} = \frac{5.00 \, \text{moles}}{2.00 \, \text{mol dm}^{-3}} = 2.50 \, \text{dm}^3
\]

#### Step 2: Convert the volume to cm³ if needed.
\[
1 \, \text{dm}^3 = 1000 \, \text{cm}^3
\]
\[
2.50 \, \text{dm}^3 = 2.50 \times 1000 \, \text{cm}^3 = 2500 \, \text{cm}^3
\]

#### Final Answer:
\[
\boxed{2.50 \, \text{dm}^3 \, \text{(or 2500 cm}^3\text{)}}
\]

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Summary of Answers:

1. Problem 4: \(\boxed{5.00 \, \text{mol dm}^{-3}}\)
2. Problem 5: \(\boxed{2.50 \, \text{mol dm}^{-3}}\)
3. Problem 6: \(\boxed{0.500 \, \text{mol dm}^{-3}}\)
4. Problem 7: \(\boxed{2.50 \, \text{dm}^3 \, \text{(or 2500 cm}^3\text{)}}\)