Molarity Problems Worksheet - Fill and Sign Printable Template Online - Free Printable

It looks like you've uploaded a Molarity Worksheet #1 with several problems related to molarity calculations. I'll solve each problem step-by-step and explain the solutions clearly.

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Problem 1: Calculate the molarity (M) of the following:

#### a. 25.0 grams of NaOH, CaCl₂, in enough water to make 450 mL of solution.
> *(Note: The problem says "NaOH, CaCl₂" — this seems like a typo or confusion. It's likely meant to be either NaOH or CaCl₂. Since both are listed, but only one mass is given, let’s assume it's a typo and it should be just one compound. Given the context, we’ll proceed with NaOH as it's more common in such problems.)*

But wait — the problem says 25.0 grams of NaOH, CaCl₂, which is ambiguous. However, looking at the answer provided in the image:
> Answer: 1.40 M

Let’s check if that matches NaOH:

- Molar mass of NaOH = 23 + 16 + 1 = 40.0 g/mol
- Moles of NaOH = \( \frac{25.0\ \text{g}}{40.0\ \text{g/mol}} = 0.625\ \text{mol} \)
- Volume = 450 mL = 0.450 L
- Molarity = \( \frac{0.625\ \text{mol}}{0.450\ \text{L}} = 1.389\ \text{M} \approx 1.39\ \text{M} \)

But the answer says 1.40 M — close enough (likely rounding).

However, the worksheet shows:
> H = 1.008, O = 16.00, Na = 22.99 → 40.0 g/mol

So yes, NaOH is intended.

✔ Corrected: 25.0 g NaOH → 0.625 mol → 1.39 M ≈ 1.40 M

✔ Answer: 1.40 M

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#### b. 75.0 grams of ethanol, CH₃CH₂OH, in 450 mL of solution.

We need to calculate molarity of ethanol.

- Molar mass of ethanol (C₂H₅OH):
- C: 12.01 × 2 = 24.02
- H: 1.008 × 6 = 6.048
- O: 16.00 × 1 = 16.00
- Total = 46.07 g/mol

- Moles = \( \frac{75.0\ \text{g}}{46.07\ \text{g/mol}} = 1.628\ \text{mol} \)

- Volume = 450 mL = 0.450 L

- Molarity = \( \frac{1.628}{0.450} = 3.62\ \text{M} \)

But the answer on the worksheet says 0.601 M — that’s way off.

Wait — did they misread the question?

Let me double-check:

> "75.0 grams of ethanol... in 450 mL"

But 75.0 g / 46.07 g/mol = ~1.63 mol → 1.63 / 0.450 = ~3.62 M

But the worksheet says:
> 0.601 M — this is wrong.

Wait — maybe they used 450 mL as 0.450 L, but perhaps they made a calculation error.

Alternatively, could it be a different substance?

No — ethanol is correct.

Wait — look at the worksheet:

> M = moles / L = 0.270 mol / 0.450 L = 0.601 M

Ah! They used 0.270 mol, not 75.0 g.

So where did 0.270 mol come from?

75.0 g / 46.07 g/mol = 1.628 mol, not 0.270.

Unless... the mass was 12.2 g? Let's reverse-calculate:

0.270 mol × 46.07 g/mol = 12.44 g, not 75.0.

So something is wrong here.

But the worksheet says:
> 75.0 g of ethanol → 0.270 mol? No.

Wait — maybe it's not ethanol, or there's a typo.

Wait — look again:
The student wrote:
> 75.0 g ethanol → 0.270 mol → 0.601 M

But that can't be right.

Let’s recalculate properly:

- Moles = 75.0 / 46.07 ≈ 1.628 mol
- M = 1.628 / 0.450 = 3.62 M

But the worksheet says 0.601 M — so unless the mass is 12.2 g, it's incorrect.

Wait — perhaps the mass is 12.2 g, not 75.0 g?

But it clearly says 75.0 g.

This appears to be an error in the worksheet or the student's work.

✔ Correct Answer:
- Moles = 75.0 / 46.07 ≈ 1.628 mol
- M = 1.628 / 0.450 = 3.62 M

✘ But the student wrote 0.601 M — this is incorrect.

So the correct molarity is ~3.62 M, not 0.601 M.

But wait — perhaps the compound is not ethanol? Or maybe it's ethanol in a different context?

No — it says CH₃CH₂OH, ethanol.

So unless there’s a typo in the mass, the student’s answer is wrong.

Let’s move on.

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#### c. 2.60 grams of sodium chloride in 40.0 mL of solution

- Molar mass of NaCl = 22.99 + 35.45 = 58.44 g/mol
- Moles = \( \frac{2.60}{58.44} = 0.0445\ \text{mol} \)
- Volume = 40.0 mL = 0.0400 L
- Molarity = \( \frac{0.0445}{0.0400} = 1.11\ \text{M} \)

But the worksheet says:
> 0.0445 mol / 0.0400 L = 1.11 M

✔ Yes, this is correct.

Answer: 1.11 M

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Problem 2: Calculate the number of liters of solution necessary to prepare:

#### a. 0.1 moles of sodium bromide to make a 0.25 M solution

Use:
\( M = \frac{\text{moles}}{\text{L}} \Rightarrow L = \frac{\text{moles}}{M} \)

\( L = \frac{0.1}{0.25} = 0.4\ \text{L} \)

Worksheet says: 0.40 L → ✔ Correct

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#### b. 3.00 moles of sodium bromide to make a 0.10 M solution

\( L = \frac{3.00}{0.10} = 30.0\ \text{L} \)

Worksheet says: 30.0 L → ✔ Correct

But then it says:
> Why did Miss Scott add extra 0 after the 3?

Because 30.0 L has three significant figures (from 3.00 moles and 0.10 M), so the answer must reflect precision.

- 3.00 has 3 sig figs
- 0.10 has 2 sig figs → limiting
- So answer should have 2 sig figs: 30 L

But the answer is written as 30.0 L — that’s 3 sig figs, which is too many.

So why did she add the extra zero?

Possibly because 3.00 ÷ 0.10 = 30.0, and students often write it that way.

But strictly speaking, 0.10 has 2 sig figs, so answer should be 30 L (2 sig figs).

So the extra zero might be incorrect for sig figs.

✔ Answer: 30 L (with 2 sig figs), but 30.0 L is acceptable if rounded up.

But technically, 30 L is better.

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Problem 3: How many moles of solute are required to prepare:

#### a. 500 mL of a 1.00 M sodium hydroxide solution

- M = 1.00 M
- V = 500 mL = 0.500 L
- Moles = M × L = 1.00 × 0.500 = 0.500 mol

Worksheet says: 0.500 mol → ✔ Correct

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#### b. 0.25 liters of a 0.008 M calcium chloride solution

- Moles = M × L = 0.008 × 0.25 = 0.0020 mol

Worksheet says: 0.0020 mol → ✔ Correct

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1a | 1.40 M (NaOH) |
| 1b | 3.62 M (Ethanol) — ✘ Student's 0.601 M is incorrect |
| 1c | 1.11 M (NaCl) |
| 2a | 0.40 L |
| 2b | 30.0 L (but should be 30 L for sig figs) |
| 3a | 0.500 mol |
| 3b | 0.0020 mol |

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🔍 Explanation of Key Concepts:

- Molarity (M) = moles of solute / liters of solution
- Always convert mL → L (divide by 1000)
- Use molar mass to convert grams → moles
- Pay attention to significant figures

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❗ Error in Problem 1b:

The student calculated:
> 75.0 g ethanol → 0.270 mol → 0.601 M

But:
- 75.0 g / 46.07 g/mol = 1.628 mol, not 0.270 mol
- So 0.601 M is wrong

✔ Correct molarity: 3.62 M

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✔ Final Correction:

Problem 1b:
- Moles = 75.0 / 46.07 ≈ 1.628 mol
- Volume = 0.450 L
- M = 1.628 / 0.450 = 3.62 M

So the student's answer is incorrect.

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