Calculating Molarity, Moles, and Volume: A Chemistry Worksheet - Free Printable

Let’s solve each problem step by step. We’ll use the molar mass of CaBr₂ to convert between moles and grams, and Avogadro’s number (6.022 × 10²³) to convert between moles and formula units.

First, find the molar mass of CaBr₂:
- Calcium (Ca): 40.08 g/mol
- Bromine (Br): 79.90 g/mol → so Br₂ = 2 × 79.90 = 159.80 g/mol
→ Molar mass of CaBr₂ = 40.08 + 159.80 = 199.88 g/mol

We’ll round to 200 g/mol for simplicity unless high precision is needed — but since the problems give values like “3.45 mol” or “267.35 grams”, we should use more precise calculation. Let’s stick with 199.88 g/mol.

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Problem 1: How many moles are in 3.45 grams of calcium bromide?

Use:
moles = mass / molar mass
= 3.45 g / 199.88 g/mol
≈ 0.01726 mol

Check: 3.45 ÷ 199.88 ≈ ?

Do division:
199.88 × 0.017 = 3.39796 → close to 3.45
Difference: 3.45 - 3.39796 = 0.05204
0.05204 / 199.88 ≈ 0.00026 → so total ≈ 0.01726 mol

✔ Answer: 0.0173 mol (rounded to 3 significant figures)

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Problem 2: Calculate the number of moles of CaBr₂ that are in 185.00 grams.

moles = 185.00 / 199.88 ≈ ?

Calculate:
199.88 × 0.925 = 199.88 × 0.9 = 179.892; 199.88 × 0.025 = 4.997 → total 184.889 → very close to 185.00

So 185.00 / 199.88 ≈ 0.9255 mol

More precisely:
185.00 ÷ 199.88 = ?

Use calculator-style:
185 ÷ 199.88 ≈ 0.92555...

✔ Answer: 0.9256 mol (rounded to 4 sig figs — since 185.00 has 5, but molar mass has 5 too, so keep 4 or 5? Problem says “185.00” — 5 sig figs. But molar mass of CaBr₂ is typically known to 4 or 5. Let’s go with 0.9256 mol.)

Actually, let’s compute exactly:

185.00 / 199.88 = 0.925555... → rounds to 0.9256 mol

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Problem 3: If a 107 gram sample contained 607.325 grams of CaBr₂, how many moles?

Wait — this seems odd. A 107 gram sample containing 607.325 grams? That can’t be right — you can’t have more mass inside than the sample itself.

Probably typo. Likely meant: “a sample contains 607.325 grams of CaBr₂” — ignore “107 gram sample”. Or maybe it’s “if a sample weighing 607.325 grams...”

Looking at context: probably just “how many moles in 607.325 grams of CaBr₂?”

Assume that.

moles = 607.325 / 199.88 ≈ ?

Compute:

199.88 × 3 = 599.64
607.325 - 599.64 = 7.685
7.685 / 199.88 ≈ 0.03845
Total ≈ 3.03845 mol

More accurately:

607.325 ÷ 199.88 = ?

Let me do: 199.88 × 3.038 = 199.88×3 = 599.64; 199.88×0.038 = ~7.595 → total 607.235 → close

Difference: 607.325 - 607.235 = 0.09 → 0.09 / 199.88 ≈ 0.00045 → so 3.03845

✔ Answer: 3.038 mol (to 4 sig figs — 607.325 has 6, but molar mass limits us? Actually, 199.88 has 5 sig figs, so answer can have 5? But let’s match input: 607.325 has 6 digits after decimal? No — 607.325 is 6 sig figs. But perhaps report as 3.0385 mol? Wait, let's calculate properly.)

Actually:

607.325 ÷ 199.88 = 3.03845... → round to 3.0385 mol? But let’s see what makes sense.

Perhaps the “107 gram sample” is a red herring or misprint. Since 607.325 > 107, likely ignore “107 gram sample” and assume it’s asking for moles in 607.325 g.

I’ll go with 3.038 mol (4 sig figs) — but let’s check exact value.

Better: 607.325 / 199.88 = 3.03845 → if we take 199.88 as exact enough, then 607.325 has 6 sig figs, so answer should have 6? But molar mass isn't infinitely precise. In practice, we'll use 3.038 mol.

Wait — let’s use calculator precision:

607.325 ÷ 199.88 = 3.038450... → so 3.0385 mol if rounded to 5 sig figs.

But looking at other problems, they use 3 or 4 sig figs. I’ll go with 3.038 mol for consistency.

Actually, let’s recalculate all with consistent precision.

Set molar mass = 199.9 g/mol (rounding to one decimal for ease? But better to use 199.88).

I think for accuracy, we’ll keep using 199.88.

Final decision: 3.038 mol

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Problem 4: How many grams of CaBr₂ are present in 0.41 mol?

mass = moles × molar mass
= 0.41 × 199.88
= ?

0.4 × 199.88 = 79.952
0.01 × 199.88 = 1.9988
Total = 81.9508 g

But 0.41 has 2 sig figs → so answer should have 2 sig figs.

81.9508 → rounds to 82 g

✔ Answer: 82 g

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Problem 5: How many moles of CaBr₂ are present in 1.05 liters of 0.50 M solution?

Molarity (M) = moles / liters → so moles = M × V

moles = 0.50 mol/L × 1.05 L = 0.525 mol

Sig figs: 0.50 has 2, 1.05 has 3 → limiting is 2 sig figs? But 0.50 might be considered 2, 1.05 is 3 — product should have 2 sig figs.

0.525 → rounds to 0.53 mol? Wait, 0.50 × 1.05 = 0.525 — which is 0.53 if rounded to 2 sig figs? But 0.525 is closer to 0.53? Actually, 0.525 to 2 sig figs is 0.53? Sig fig rules for multiplication: least number of sig figs.

0.50 has 2 sig figs, 1.05 has 3 → so answer has 2 sig figs.

0.525 → the first two digits are 5 and 2, third is 5 → round up → 0.53

But 0.53 has two sig figs? Yes.

Some might argue 0.50 is ambiguous, but typically treated as 2 sig figs.

✔ Answer: 0.53 mol

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Problem 6: How many liters of 0.100 M calcium bromide contain 1.00 mole of CaBr₂?

M = moles / volume → volume = moles / M

volume = 1.00 mol / 0.100 mol/L = 10.0 L

Sig figs: 1.00 has 3, 0.100 has 3 → answer 10.0 L (3 sig figs)

✔ Answer: 10.0 L

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Problem 7: What volume of 0.100 M solution would contain 0.050 moles of CaBr₂?

volume = moles / M = 0.050 / 0.100 = 0.50 L

Sig figs: both have 2 sig figs → 0.50 L

✔ Answer: 0.50 L

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Problem 8: How many grams of CaBr₂ are needed to prepare 1.00 liter of 0.500 M solution?

First, moles needed = M × V = 0.500 mol/L × 1.00 L = 0.500 mol

Then mass = moles × molar mass = 0.500 × 199.88 = 99.94 g

Sig figs: 0.500 has 3, 1.00 has 3, molar mass 199.88 has 5 → so answer 3 sig figs.

99.94 → rounds to 99.9 g

But 0.500 × 199.88 = 99.94 → yes.

Sometimes written as 100. g if rounded, but 99.94 is closer to 99.9.

Wait: 0.500 × 199.88 = let's compute:

0.5 × 199.88 = 99.94 → exactly.

With 3 sig figs: 99.9 g

But 99.94 rounded to 3 sig figs is 99.9? Yes.

However, sometimes in chemistry, if it's 99.94 and you need 3 sig figs, it's 99.9.

But let's confirm: 99.94 — the third digit is 9, fourth is 4 <5, so 99.9 g.

✔ Answer: 99.9 g

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Now the extra credit part:

“Use what the previous lesson covered about in a real-world scenario: Buckyballs (C₆₀) are molecules made of 60 carbon atoms.”

Given: atomic mass of C = 12.01 amu → so molar mass of C₆₀ = 60 × 12.01 = 720.6 g/mol

Avogadro’s number: 6.022 × 10²³ molecules/mol

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Extra Credit Q1: How many grams of C₆₀ are needed to make 1.00 mole?

That’s just the molar mass: 720.6 g

✔ Answer: 720.6 g

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Extra Credit Q2: How many molecules of C₆₀ are needed to make 1.00 mole?

By definition, 1 mole = 6.022 × 10²³ molecules

✔ Answer: 6.022 × 10²³ molecules

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Extra Credit Q3: How many grams of C₆₀ are needed to make 1.00 × 10²⁴ molecules?

First, find moles: moles = number of molecules / Avogadro’s number

= 1.00 × 10²⁴ / 6.022 × 10²³ = 1.6606 mol

Then mass = moles × molar mass = 1.6606 × 720.6 ≈ ?

Compute:

1.6606 × 700 = 1162.42
1.6606 × 20.6 = 34.20836
Total ≈ 1196.62836 g

Sig figs: 1.00 × 10²⁴ has 3 sig figs, Avogadro’s number is exact for this purpose? Usually taken as 6.022 × 10²³ with 4 sig figs, but often treated as exact in such contexts. Molar mass 720.6 has 4 sig figs.

Input: 1.00 × 10²⁴ → 3 sig figs → so answer should have 3 sig figs.

1196.62836 → rounds to 1200 g? But that’s 2 sig figs if written as 1200. Better to write as 1.20 × 10³ g

Compute exactly:

1.00e24 / 6.022e23 = 1.660577... mol

1.660577 × 720.6 = let's calculate:

1.660577 × 700 = 1162.4039
1.660577 × 20.6 = 34.2078862
Sum = 1196.6117862 g

To 3 sig figs: 1.20 × 10³ g

✔ Answer: 1.20 × 10³ g or 1200 g (but specify as 1.20 × 10³ to show 3 sig figs)

In boxed answer, we can write 1200 g if context allows, but better 1.20 \times 10^3 g.

Since the problem says "grams", and doesn't specify format, but to be precise, I'll use 1200 g with note, but in final answer, perhaps 1.20 \times 10^3.

Looking at other answers, they use decimals, so for consistency, maybe 1200 g is acceptable, but technically 1200 has ambiguous sig figs.

I think for this level, 1200 g is fine, implying 3 sig figs from context.

But let's see: 1.00 × 10²⁴ has 3 sig figs, so answer must have 3. 1200 written without decimal is ambiguous. So best to write as 1.20 × 10³ g

In text, we can say 1200 g, but in final answer box, perhaps write 1200.

I recall that in many textbooks, they accept 1200 g for 3 sig figs when it's clear.

To be safe, I'll calculate and round properly.

1196.6 g → to 3 sig figs: look at first three digits: 1,1,9 — next is 6>5, so round up 119 to 120 → so 1200 g, which is 1.20 × 10³ g.

Yes.

✔ Answer: 1.20 × 10³ g

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Now, compiling all answers:

1. 0.0173 mol
2. 0.9256 mol
3. 3.038 mol
4. 82 g
5. 0.53 mol
6. 10.0 L
7. 0.50 L
8. 99.9 g

Extra Credit:
1. 720.6 g
2. 6.022 × 10²³ molecules
3. 1.20 × 10³ g

But for problem 3, I was unsure about the "107 gram sample". Re-reading: "If a 107 gram sample contained 607.325 grams of CaBr₂" — this is impossible. Probably a typo, and it should be "if a sample contains 607.325 grams" or "if 607.325 grams of CaBr₂ are in a sample". I think it's safe to assume it's asking for moles in 607.325 g.

Perhaps "107" is a label or something else. I'll proceed with 607.325 g.

Also, for problem 2, 185.00 / 199.88 = let's compute exactly:

185.00 ÷ 199.88 = 0.925555... → 0.9256 mol (since 185.00 has 5 sig figs, molar mass 199.88 has 5, so 0.9256 has 4? Wait, 0.9256 has 4 sig figs, but it's 9.256 × 10^{-1}, so 4 sig figs. Input 185.00 has 5, but molar mass might be limiting. Typically, we take the least precise, but 199.88 is usually considered precise. I think 0.9256 is fine.

Similarly, for problem 1: 3.45 / 199.88 = 0.01726 → 0.0173 mol (3 sig figs)

Now, for the final answer section, since there are multiple parts, I need to list them all.

The user didn't specify to separate, but since it's a worksheet, probably list each answer.

In the response format, it says "Final Answer: [Provide only the final correct answer]", but for multiple questions, I should provide all.

Looking back at instructions: "Solve the problem accurately." and the image has multiple problems.

I think I need to provide answers for all.

Perhaps box each or list.

Since it's a single response, I'll list them in order.

For clarity, I'll number them.

Final Answer:

1. 0.0173 mol
2. 0.9256 mol
3. 3.038 mol
4. 82 g
5. 0.53 mol
6. 10.0 L
7. 0.50 L
8. 99.9 g
Extra Credit 1: 720.6 g
Extra Credit 2: 6.022 × 10²³ molecules
Extra Credit 3: 1.20 × 10³ g

But for Extra Credit 3, perhaps write as 1200 g if allowed, but I'll keep scientific notation.

To match the style, maybe write all in decimal where possible.

For EC3, 1200 g is acceptable.

I think for school level, 1200 g is fine.

Let me confirm calculation again for EC3:

Number of molecules = 1.00 × 10²⁴

Moles = 1.00e24 / 6.022e23 = 1.660577... mol

Molar mass C60 = 60 * 12.01 = 720.6 g/mol

Mass = 1.660577 * 720.6 = let's compute:

1.660577 * 720.6 =

First, 1.660577 * 700 = 1162.4039

1.660577 * 20.6 = 34.2078862

Sum: 1162.4039 + 34.2078862 = 1196.6117862 g

Rounded to 3 significant figures: 1.20 × 10³ g or 1200 g (with the understanding that it's 3 sig figs)

In many contexts, they write 1200. g with a decimal to indicate 4 sig figs, but here it's 3, so 1.20 × 10³ is best.

But for simplicity, since other answers are in decimal, and 1200 is common, I'll use 1200 g, assuming it's understood as 3 sig figs from context.

To be precise, I'll use 1.20 \times 10^3 g in the answer.

In text, I can write "1200 g (1.20 × 10³ g)" but for final answer, perhaps just 1200 g.

I think for this platform, writing 1200 g is acceptable.

Let's see the other answers: problem 4 is 82 g, which is 2 sig figs, so 1200 g for 3 sig figs is ok.

So I'll go with 1200 g for EC3.

Similarly, for problem 3, 3.038 mol is fine.

Now, one more thing: in problem 5, I said 0.53 mol, but let's double-check sig figs.

0.50 M has 2 sig figs, 1.05 L has 3 sig figs. The product should have 2 sig figs.

0.50 * 1.05 = 0.525, which rounds to 0.53 (since 0.525, the third digit is 5, so round up the 2 to 3? 0.525 to 2 sig figs: the first non-zero digit is 5, second is 2, third is 5, so round up 2 to 3, so 0.53.

Yes.

Some might argue that 0.50 has uncertainty in the hundredths place, but standard rule is 2 sig figs.

So 0.53 mol is correct.

Now, for the final answer listing.

I think I have all.

So Final Answer will be a list.

Since the instruction says "provide only the final correct answer", and it's multiple, I'll list them clearly.

Perhaps in a boxed format, but since it's text, I'll write:

Final Answer:
1. 0.0173 mol
2. 0.9256 mol
3. 3.038 mol
4. 82 g
5. 0.53 mol
6. 10.0 L
7. 0.50 L
8. 99.9 g
Extra Credit 1: 720.6 g
Extra Credit 2: 6.022 × 10²³ molecules
Extra Credit 3: 1200 g

For EC3, to be accurate, I'll use 1.20 × 10³ g, but I think 1200 g is fine.

Upon second thought, in the calculation, 1196.6 g, to 3 sig figs is 1.20 × 10³ g, which is 1200 g, but written as 1200 implies 4 sig figs if no decimal, but in context, it's ok.

To avoid confusion, I'll write it as 1.20 \times 10^3 g in the answer.

But in plain text, "1.20e3 g" or "1200 g".

I think for this, "1200 g" is acceptable.

Let's move on.

One last check on problem 3: if the "107 gram sample" is important, perhaps it's a mixture, and 607.325 g is not possible, so likely typo. Maybe it's "60.7325 g" or something, but as given, I'll assume 607.325 g.

Perhaps "107" is the problem number or something, but in the text, it's "If a 107 gram sample contained 607.325 grams" — definitely error. I think it's safe to ignore "107 gram sample" and take 607.325 g as the mass of CaBr2.

So I'll proceed.

Final Answer compilation.