Molarity Worksheet #1 for chemistry practice, including calculations for molarity, moles, and solution preparation.

Let’s solve each problem step by step. We’ll use the molarity formula:

Molarity (M) = moles of solute / liters of solution

We’ll also remember:
- To convert mL to L → divide by 1000
- To find moles from grams → divide mass by molar mass
- Molar mass is found by adding atomic masses from the periodic table

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Problem 1a:

75.0 grams of NaCl in enough water to make 450. mL of solution

Step 1: Find molar mass of NaCl
Na = 23.0 g/mol, Cl = 35.5 g/mol → NaCl = 23.0 + 35.5 = 58.5 g/mol

Step 2: Convert grams to moles
moles = 75.0 g ÷ 58.5 g/mol ≈ 1.282 mol

Step 3: Convert mL to L
450. mL = 450. ÷ 1000 = 0.450 L

Step 4: Calculate molarity
M = 1.282 mol ÷ 0.450 L ≈ 2.85 M

✔ Final Answer for 1a: 2.85 M

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Problem 1b:

75.0 grams of ethane, C₂H₆, in 450. mL of solution

Step 1: Molar mass of C₂H₆
C = 12.0 × 2 = 24.0
H = 1.0 × 6 = 6.0
Total = 24.0 + 6.0 = 30.0 g/mol

Step 2: Moles of C₂H₆
75.0 g ÷ 30.0 g/mol = 2.50 mol

Step 3: Volume in L = 0.450 L (same as above)

Step 4: Molarity = 2.50 mol ÷ 0.450 L ≈ 5.56 M

✔ Final Answer for 1b: 5.56 M

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Problem 1c:

2.60 grams of sodium chloride (NaCl) in 40.0 mL of solution

Wait — this says “sodium chloride” again? But it gives a different mass and volume. Let’s check if it’s really NaCl or maybe something else? The original image shows “LiCl” in the answer key, but the question says “sodium chloride”. That might be a typo.

Looking at the handwritten answer: they used LiCl (lithium chloride), not NaCl.

Let’s assume it’s LiCl, since that matches the answer shown.

Molar mass of LiCl:
Li = 6.9, Cl = 35.5 → 6.9 + 35.5 = 42.4 g/mol

Moles = 2.60 g ÷ 42.4 g/mol ≈ 0.0613 mol

Volume = 40.0 mL = 0.0400 L

Molarity = 0.0613 mol ÷ 0.0400 L ≈ 1.53 M

But wait — the handwritten answer says 0.144 M. That doesn’t match.

Hold on — let’s recalculate with exact numbers.

If we use Li = 6.94, Cl = 35.45 → LiCl = 42.39 g/mol

2.60 ÷ 42.39 ≈ 0.06134 mol

0.06134 ÷ 0.0400 = 1.5335 M → still ~1.53 M

But the image shows 0.144 M. That suggests maybe it’s not LiCl?

Wait — perhaps it’s NaCl after all? Let’s try that.

NaCl = 58.5 g/mol

2.60 g ÷ 58.5 g/mol ≈ 0.04444 mol

÷ 0.0400 L = 1.111 M — still not 0.144

Hmm. What if it’s KCl? K = 39.1, Cl = 35.5 → 74.6 g/mol

2.60 ÷ 74.6 ≈ 0.03485 mol ÷ 0.0400 = 0.871 M — no.

What if it’s CaCl₂? Ca=40, Cl₂=71 → 111 g/mol

2.60 ÷ 111 ≈ 0.0234 mol ÷ 0.0400 = 0.585 M — no.

Wait — look at the handwritten work:

They wrote:
“2.60 g LiCl × 1 mol / 42.45g = 0.0613 mol”
Then “0.0613 mol / 0.040 L = 1.53 M” — but then they crossed out and wrote 0.144 M? That doesn’t make sense.

Actually, looking closely — they have:

> 2.60 g LiCl × 1 mol / 42.45g = 0.0613 mol
> Then: 0.0613 mol / 0.040 L = 1.53 M — but then they wrote “= 0.144 M” ??? That must be a mistake.

Wait — unless... did they mean 2.60 mg? No, it says grams.

Alternatively — maybe the compound is AlCl₃? Al=27, Cl₃=106.5 → 133.5 g/mol

2.60 ÷ 133.5 ≈ 0.01948 mol ÷ 0.0400 = 0.487 M — no.

I think there’s an error in the image’s handwritten answer. Based on correct calculation for LiCl, it should be 1.53 M.

But let’s double-check the problem statement: “2.60 grams of sodium chloride” — so it says sodium chloride, which is NaCl.

So let’s do it correctly for NaCl:

Molar mass NaCl = 58.44 g/mol (more precise)

2.60 g ÷ 58.44 g/mol = 0.04449 mol

Volume = 40.0 mL = 0.0400 L

M = 0.04449 ÷ 0.0400 = 1.112 M

Still not 0.144.

Wait — what if the volume is 400 mL? No, it says 40.0 mL.

Another possibility: maybe it’s 2.60 grams of LiCl in 400 mL? But it says 40.0 mL.

I think the handwritten answer has a calculation error. Let’s stick to correct math.

But to match the image’s final answer of 0.144 M, let’s reverse-engineer:

If M = 0.144 mol/L, V = 0.040 L → moles = 0.144 × 0.040 = 0.00576 mol

Mass = 2.60 g → molar mass = 2.60 / 0.00576 ≈ 451 g/mol — that’s way too high for any common salt.

This suggests a possible typo in the problem or answer.

Given the confusion, and since the problem says “sodium chloride”, I’ll go with NaCl.

Final calculation for 1c (assuming NaCl):

Molar mass NaCl = 58.44 g/mol
moles = 2.60 / 58.44 = 0.04449 mol
volume = 0.0400 L
M = 0.04449 / 0.0400 = 1.11 M

But the image shows 0.144 M — which is likely wrong.

However, looking back at the image, in part c, they wrote:

> 2.60 g LiCl × 1 mol / 42.45g = 0.0613 mol
> Then: 0.0613 mol / 0.040 L = 1.53 M — but then they have "= 0.144 M" written below? That seems like a mistake.

Actually, in the image, under 1c, they have:

> = 0.144 M

And before that, they have:

> 2.60 g LiCl × 1 mol / 42.45g = 0.0613 mol
> 0.0613 mol / 0.040 L = ?

0.0613 / 0.040 = 1.5325 — not 0.144.

Unless they meant 0.0613 / 0.425 or something? No.

Perhaps they divided by 0.425 L by mistake? 0.0613 / 0.425 = 0.144 — ah! Maybe they used 425 mL instead of 40.0 mL?

That would explain it. If volume was 425 mL = 0.425 L, then 0.0613 / 0.425 = 0.144 M.

But the problem says 40.0 mL.

Given the discrepancy, and since the problem explicitly says "40.0 mL", I will calculate correctly.

For LiCl (as per their choice):
M = 2.60 g / 42.45 g/mol / 0.0400 L = 0.06125 / 0.0400 = 1.53 M

For NaCl: 2.60 / 58.44 / 0.0400 = 0.04449 / 0.0400 = 1.11 M

Since the problem says "sodium chloride", I'll go with NaCl.

But to match the context of the worksheet (which uses LiCl in the answer), and to avoid confusion, I'll note that there's an inconsistency.

However, for accuracy, let's assume the compound is as stated: sodium chloride (NaCl).

So for 1c: 1.11 M

But the image's answer is 0.144 M, which is incorrect based on given data.

Perhaps it's a different compound? Let's try MgCl2: Mg=24.3, Cl2=71, total 95.3 g/mol

2.60 / 95.3 = 0.02728 mol / 0.0400 L = 0.682 M — no.

Or AlCl3: 133.34 g/mol — 2.60/133.34=0.0195 mol /0.04=0.487 M — no.

I think the best approach is to use the compound as written: sodium chloride.

So 1c: 1.11 M

But let's move on and come back.

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Problem 2a:

How many liters of solution necessary to prepare 5.0 grams of sodium bromide from a 0.25 M solution?

First, find moles of NaBr.

Molar mass NaBr: Na=23.0, Br=79.9 → 102.9 g/mol

moles = 5.0 g / 102.9 g/mol ≈ 0.04859 mol

Molarity = moles / volume(L) → volume = moles / M

V = 0.04859 mol / 0.25 mol/L = 0.19436 L

Round to significant figures: 5.0 g has 2 sig figs, 0.25 M has 2 sig figs → answer should have 2 sig figs.

0.19436 L ≈ 0.19 L

The image shows 0.19 L — good.

Note: They wrote "Why did Miss Scott add that extra 0 after the .?" — probably referring to writing 0.19 vs 0.190, but with 2 sig figs, 0.19 is fine.

✔ Final Answer for 2a: 0.19 L

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Problem 2b:

2.00 grams of strontium nitrate from a 0.10 M solution

Strontium nitrate: Sr(NO3)2

Molar mass: Sr=87.6, N=14.0×2=28.0, O=16.0×6=96.0 → total = 87.6+28.0+96.0 = 211.6 g/mol

moles = 2.00 g / 211.6 g/mol ≈ 0.009452 mol

M = 0.10 M → V = moles / M = 0.009452 / 0.10 = 0.09452 L

Sig figs: 2.00 g has 3, 0.10 M has 2 → limit to 2 sig figs.

0.09452 L ≈ 0.095 L

Image shows 0.094 L — close, but with rounding.

0.009452 / 0.10 = 0.09452, which rounds to 0.095 if using two sig figs? Wait, 0.10 has two sig figs, so answer should have two.

0.09452 — the first two non-zero digits are 9 and 4, so 0.095 L (since 4 is followed by 5, round up).

But 0.09452 to two sig figs: look at 9.452 × 10^-2 — two sig figs is 9.5 × 10^-2 = 0.095 L.

Image has 0.094 L — perhaps they used slightly different molar mass.

If Sr=87.62, N=14.01×2=28.02, O=16.00×6=96.00 → total 211.64 g/mol

2.00 / 211.64 = 0.009449 mol

/ 0.10 = 0.09449 L → which rounds to 0.094 L if keeping three decimal places, but sig figs suggest 0.095 L.

To match common practice, and since 0.10 has two sig figs, 0.095 L is better.

But the image shows 0.094 L, so perhaps they expect that.

Let's calculate exactly as per standard values.

Use: Sr = 87.62, N = 14.01, O = 16.00

Sr(NO3)2 = 87.62 + 2*(14.01 + 48.00) = 87.62 + 2*62.01 = 87.62 + 124.02 = 211.64 g/mol

moles = 2.00 / 211.64 = 0.009449 mol

V = 0.009449 / 0.10 = 0.09449 L

With two significant figures (from 0.10 M), it should be 0.094 L? No — 0.10 has two sig figs, but it's 1.0 × 10^{-1}, so when dividing, the result should have two sig figs.

0.09449 L — the first two significant digits are 9 and 4, and the next digit is 4, which is less than 5, so it should be 0.094 L.

Yes, because 0.09449 rounded to two sig figs is 0.094 L.

How? 9.449 × 10^{-2} — two sig figs: look at the third digit, which is 4 < 5, so 9.4 × 10^{-2} = 0.094 L.

Yes.

✔ Final Answer for 2b: 0.094 L

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Problem 3a:

How many moles of substances are required to prepare 500 mL of a 0.100 M hydrochloric acid solution?

Molarity = moles / volume(L)

So moles = M × V(L)

V = 500 mL = 0.500 L

M = 0.100 mol/L

moles = 0.100 × 0.500 = 0.0500 mol

Image shows 0.0500 mol — correct.

✔ Final Answer for 3a: 0.0500 mol

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Problem 3b:

0.25 liters of a 0.00 M calcium chloride solution

Wait — 0.00 M? That can't be right. Probably a typo.

In the image, it says "0.00 M", but then they calculate with 0.500 M.

Look: "moles = 0.500 M x 0.25 L = 0.125 mol CaCl2"

And the problem says "0.00 M" — likely a misprint; should be 0.500 M.

Because 0.00 M would give 0 moles, which is trivial.

In the image, they have "0.500 M" in the calculation.

So I'll assume it's 0.500 M.

moles = M × V = 0.500 mol/L × 0.25 L = 0.125 mol

Sig figs: 0.500 has 3, 0.25 has 2 → answer should have 2 sig figs.

0.125 rounded to two sig figs is 0.13 mol? But 0.25 has two sig figs, 0.500 has three, so product has two.

0.125 — the '1' and '2' are the first two digits, but since it's 1.25 × 10^{-1}, two sig figs would be 1.3 × 10^{-1} = 0.13 mol.

But the image shows 0.125 mol, and they didn't round.

Typically in such problems, if volumes are given as 0.25 L (which might be considered as having two sig figs), but sometimes it's exact.

To be precise, 0.25 L could be 250 mL, which might be measured to nearest 10 mL, so two sig figs.

But in many textbooks, they keep it as is.

Given that the image has 0.125 mol, and it's a direct multiplication, I'll go with that.

But technically, with sig figs, it should be 0.13 mol.

However, let's see the problem: "0.25 liters" — if it's exactly 0.25, it might be considered as having infinite precision, but usually in chemistry, we take the digits given.

0.25 has two sig figs, 0.500 has three, so product has two.

0.500 × 0.25 = 0.125 → which is 1.25 × 10^{-1}, so to two sig figs, 1.3 × 10^{-1} = 0.13 mol.

But the image has 0.125, so perhaps they want the exact value.

To match the context, I'll put 0.125 mol, noting that sig figs might require 0.13.

But let's check the image: they have "0.125 mol CaCl2" — so I'll go with that.

✔ Final Answer for 3b: 0.125 mol

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Now, back to 1c. Given the confusion, and since the problem says "sodium chloride", I'll calculate for NaCl.

But to resolve, let's assume the compound is as written: sodium chloride (NaCl).

So for 1c:
mass = 2.60 g NaCl
molar mass = 58.44 g/mol
moles = 2.60 / 58.44 = 0.04449 mol
volume = 40.0 mL = 0.0400 L
M = 0.04449 / 0.0400 = 1.11225 M → 1.11 M (three sig figs)

But the image has 0.144 M, which is likely for a different compound or volume.

Perhaps it's 2.60 grams in 400 mL? Let's try:

If volume = 400 mL = 0.400 L
M = 0.04449 / 0.400 = 0.111 M — still not 0.144.

If it's LiCl in 400 mL: moles = 2.60 / 42.45 = 0.06125 mol
V = 0.400 L
M = 0.06125 / 0.400 = 0.153 M — close to 0.144? Not really.

2.60 / 42.45 = 0.06125, / 0.425 = 0.144 — so if volume was 425 mL, it would be 0.144 M.

But the problem says 40.0 mL.

I think there's a typo in the problem or answer key. For the sake of this exercise, I'll use the correct calculation based on given data.

Since the problem says "sodium chloride", and 40.0 mL, I'll go with 1.11 M.

But to match the worksheet's intent, and since they used LiCl in the answer, perhaps it's LiCl.

Let me calculate for LiCl with 40.0 mL:

Molar mass LiCl = 6.94 + 35.45 = 42.39 g/mol
moles = 2.60 / 42.39 = 0.06134 mol
V = 0.0400 L
M = 0.06134 / 0.0400 = 1.5335 M → 1.53 M

The image has 0.144 M, which is incorrect.

Perhaps the mass is 0.260 g? 0.260 / 42.39 = 0.006134 mol / 0.0400 = 0.153 M — still not 0.144.

Or 0.244 g / 42.39 = 0.005756 mol / 0.0400 = 0.1439 M ≈ 0.144 M.

So if mass was 0.244 g, it would work.

But the problem says 2.60 g.

I think the best course is to report the correct calculation for the given numbers.

For 1c, assuming sodium chloride (NaCl): 1.11 M

But to align with the worksheet's answer key (which shows 0.144 M for LiCl), and since the problem might have a typo, I'll provide both, but for the final answer, I'll use the calculation as per standard.

Perhaps in the worksheet, it's "2.60 grams of lithium chloride" — let's assume that, as per their calculation.

In the image, they have "LiCl" in the work, so likely the problem meant lithium chloride.

So for 1c: lithium chloride, 2.60 g in 40.0 mL.

Molar mass LiCl = 42.39 g/mol (standard)

moles = 2.60 / 42.39 = 0.06134 mol

V = 0.0400 L

M = 0.06134 / 0.0400 = 1.5335 M → 1.53 M

But their answer is 0.144 M, which is wrong.

Unless... in the image, for 1c, they have:

> 2.60 g LiCl × 1 mol / 42.45g = 0.0613 mol
> Then: 0.0613 mol / 0.040 L = ?

And then they have "= 0.144 M" — but 0.0613 / 0.040 = 1.5325, not 0.144.

Perhaps they meant to write 0.0613 / 0.425 = 0.144, and the volume is 425 mL, but it's written as 40.0 mL.

I think for accuracy, I'll calculate correctly.

So for 1c, with LiCl: 1.53 M

But to match the expected answer, and since the user might be expecting the image's answer, I'll note the discrepancy.

For the purpose of this response, I'll use the correct math.

Let's finalize all answers.

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Final Answers:

1a: 2.85 M
1b: 5.56 M
1c: 1.53 M (for LiCl, as per worksheet's choice) or 1.11 M for NaCl — I'll go with 1.53 M since the worksheet used LiCl.
2a: 0.19 L
2b: 0.094 L
3a: 0.0500 mol
3b: 0.125 mol

For 1c, to be consistent with the worksheet's approach, I'll use LiCl and 1.53 M, even though the problem says "sodium chloride" — likely a typo in the problem.

So:

Final Answer:
1a: 2.85 M
1b: 5.56 M
1c: 1.53 M
2a: 0.19 L
2b: 0.094 L
3a: 0.0500 mol
3b: 0.125 mol

But let's box them as per the format.

Since the instruction is to provide only the final correct answer after reasoning, and there are multiple parts, I'll list them.

Final Answer:
1a: 2.85 M
1b: 5.56 M
1c: 1.53 M
2a: 0.19 L
2b: 0.094 L
3a: 0.0500 mol
3b: 0.125 mol

Note: For 1c, if the compound is indeed sodium chloride, it should be 1.11 M, but based on the worksheet's calculation, it's treated as LiCl yielding 1.53 M. The handwritten answer of 0.144 M appears to be incorrect for the given values.