Answers Percent Concentration Worksheet | PDF - Free Printable

You've uploaded an image showing the “ANSWERS - Percent Concentration Practice Worksheet”, which includes 6 problems involving percent concentration calculations — specifically mass/mass (% M/M) and volume/volume (% V/V).

Since you asked to “solve the problem and explain the solution,” I’ll go through each question one by one, explaining how to arrive at each answer — even though the answers are already provided in the image. This will help you understand the *reasoning* behind each calculation.

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🔢 1. What is the % M/M if 8.0 g copper is added to enough zinc to produce 100 g of an alloy?

✔ Answer: 8% M/M

#### 📌 Explanation:
% M/M = (mass of solute / mass of solution) × 100%

- Solute = copper = 8.0 g
- Solution = total alloy = 100 g

> % M/M = (8.0 g / 100 g) × 100% = 8%

This is a straightforward mass percent calculation. The “enough zinc” just means the rest of the 100 g is zinc — so total mass is 100 g.

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🔢 2. Brass is a copper-zinc alloy... If a 35.0 g pendant contains 1.7 g of zinc, what is the percentage mass by mass of zinc in brass?

✔ Answer: 4.9% M/M

#### 📌 Explanation:
Again, use:

> % M/M = (mass of component / total mass) × 100%

- Component = zinc = 1.7 g
- Total mass of brass pendant = 35.0 g

> % M/M = (1.7 g / 35.0 g) × 100% ≈ 4.857% → rounded to 4.9%

Note: They likely rounded to 1 decimal place.

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🔢 3. What is the % M/M of copper in an alloy when 10 kg of Cu is mixed with 250 kg of Zn?

✔ Answer: 3.8% M/M

#### 📌 Explanation:
Total mass of alloy = mass of Cu + mass of Zn = 10 kg + 250 kg = 260 kg

> % M/M of Cu = (10 kg / 260 kg) × 100% ≈ 3.846% → rounded to 3.8%

Important: Units must match — both in kg, so no conversion needed.

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🔢 4. You have 200 g of a solution that contains 30 g of hydrochloric acid (HCl), what percentage of your solution is made up of hydrochloric acid (M/M)?

✔ Answer: 15% M/M

#### 📌 Explanation:
> % M/M = (mass of HCl / total mass of solution) × 100%

= (30 g / 200 g) × 100% = 15%

Simple ratio — 30 is one-fifth of 200, and 1/5 = 20%? Wait — 30/200 = 0.15 → 15%. Correct.

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🔢 5. If I make a solution by adding water to 75 mL of ethanol until the total volume of the solution is 375 mL, what's the percent by volume of ethanol in the solution?

✔ Answer: 20% V/V

#### 📌 Explanation:
% V/V = (volume of solute / volume of solution) × 100%

- Solute = ethanol = 75 mL
- Solution = total volume = 375 mL

> % V/V = (75 mL / 375 mL) × 100% = 20%

Note: This assumes volumes are additive (which isn’t always true in chemistry, but for this level, we assume it).

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🔢 6. How would you make 500 mL of 30% vinegar in water solution?

✔ Answer: Add 150 mL of vinegar to water and dilute to 500 mL total volume.

#### 📌 Explanation:
This is a volume/volume percent problem.

> % V/V = (volume of solute / volume of solution) × 100%

We want 30% V/V vinegar in 500 mL total solution.

So,

> Volume of vinegar = (30 / 100) × 500 mL = 150 mL

Then, add water to reach the final volume of 500 mL.

⚠️ Important: You do NOT add 350 mL of water to 150 mL vinegar — because mixing liquids may not give exactly 500 mL due to intermolecular interactions. So, correct lab procedure is:

> Add 150 mL vinegar to a volumetric flask, then add water slowly while stirring until the total volume reaches 500 mL.

That’s why the worksheet says:
> “So, you would place 150 mL of vinegar in a flask and bring to level (500 mL) with water.”

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## ✔ Summary Table:

| Problem | Type | Calculation | Answer |
|---------|----------|--------------------------------------------------|------------|
| 1 | % M/M | (8.0 / 100) × 100% | 8% |
| 2 | % M/M | (1.7 / 35.0) × 100% | 4.9% |
| 3 | % M/M | (10 / 260) × 100% | 3.8% |
| 4 | % M/M | (30 / 200) × 100% | 15% |
| 5 | % V/V | (75 / 375) × 100% | 20% |
| 6 | % V/V prep | 30% of 500 mL = 150 mL vinegar → dilute to 500 mL | 150 mL vinegar + water to 500 mL |

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## 💡 Key Takeaways:

- % M/M = (mass of solute / mass of solution) × 100%
- % V/V = (volume of solute / volume of solution) × 100%
- Always ensure units match.
- For preparing solutions, add solute first, then dilute to final volume — don’t assume volumes are additive unless specified.

Let me know if you’d like practice problems or step-by-step worksheets!