Molarity Practice Worksheet featuring six chemistry problems to calculate molarity of different solutions.

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

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

We must make sure units are correct — especially converting mL to L when needed, and grams to moles using molar mass.

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Problem 1:

> 0.5 moles of sodium chloride is dissolved to make 0.05 liters of solution.

This is straightforward.

M = moles / liters
M = 0.5 mol / 0.05 L = 10 M

✔ Check: 0.5 ÷ 0.05 = 10 → Correct.

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Problem 2:

> 0.5 grams of sodium chloride is dissolved to make 0.05 liters of solution.

First, convert grams to moles.

Molar mass of NaCl = 23 (Na) + 35.5 (Cl) = 58.5 g/mol

Moles = mass / molar mass = 0.5 g / 58.5 g/mol ≈ 0.008547 mol

Now, M = moles / liters = 0.008547 mol / 0.05 L ≈ 0.1709 M

Round to 2 or 3 significant figures? The given values (0.5 g, 0.05 L) have 1 or 2 sig figs. Let’s go with 2 sig figs.

→ 0.17 M

✔ Check: 0.5 / 58.5 = ~0.00855; divided by 0.05 = ~0.171 → rounds to 0.17 M

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Problem 3:

> 0.5 grams of sodium chloride is dissolved to make 0.05 mL of solution.

Wait — volume is in mL, not liters! Convert to liters first.

0.05 mL = 0.05 / 1000 = 0.00005 L

Moles of NaCl = same as above: 0.5 g / 58.5 g/mol ≈ 0.008547 mol

M = 0.008547 mol / 0.00005 L = 170.94 M

That’s extremely concentrated — but mathematically correct based on tiny volume.

Sig figs: 0.5 g (1 sig fig), 0.05 mL (1 sig fig) → answer should be 200 M? Wait — let’s think.

Actually, 0.5 has 1 sig fig, 0.05 has 1 sig fig → so answer should have 1 sig fig.

170.94 → rounded to 1 sig fig = 200 M

But that feels odd. Maybe we keep it as 170 M if we consider 0.05 as having 1 sig fig but sometimes trailing zeros after decimal count... Actually, 0.05 has 1 sig fig. So yes, 200 M.

But let’s double-check calculation:

0.5 / 58.5 = 0.008547

0.008547 / 0.00005 = 170.94 → which is 1.7 × 10² → with 1 sig fig → 2 × 10² = 200 M

✔ Final: 200 M

*(Note: This is unrealistic in real life, but mathematically correct for the problem.)*

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Problem 4:

> 734 grams of lithium sulfate are dissolved to make 2500 mL of solution.

First, find molar mass of Li₂SO₄.

Li: 6.94 × 2 = 13.88
S: 32.07
O: 16.00 × 4 = 64.00
Total = 13.88 + 32.07 + 64.00 = 109.95 g/mol ≈ 110.0 g/mol (we’ll use 110 for simplicity, but let’s be precise)

Actually, better to use exact:

Li: 6.941 × 2 = 13.882
S: 32.065
O: 15.999 × 4 = 63.996
Total = 13.882 + 32.065 + 63.996 = 109.943 g/mol

Moles = 734 g / 109.943 g/mol ≈ ?

Calculate: 734 ÷ 109.943 ≈ 6.676 mol

Volume = 2500 mL = 2.500 L

M = 6.676 mol / 2.500 L ≈ 2.670 M

Sig figs: 734 g (3 sig figs), 2500 mL (could be 2, 3, or 4 — assume 4 since written as 2500. without decimal? But usually 2500 has 2 sig figs unless specified. Hmm.)

In chemistry problems, if it says “2500 mL” without a decimal, often treated as 2 sig figs. But 734 has 3. So limiting is 2 sig figs? Or maybe 3?

Actually, 2500 mL — if it’s exactly 2500, it might be considered 4 sig figs. But to be safe, let’s look at context. Since 734 has 3, and 2500 could be ambiguous, I’ll go with 3 sig figs.

So 2.67 M

✔ Check: 734 / 109.943 ≈ 6.676; / 2.5 = 2.6704 → 2.67 M

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Problem 5:

> 6.7 x 10⁻² grams of Pb(C₂H₃O₂)₄ are dissolved to make 3.5 mL of solution.

First, find molar mass of Pb(C₂H₃O₂)₄.

Breakdown:

Pb: 207.2
C₂H₃O₂ is acetate ion: C₂ = 24.02, H₃ = 3.024, O₂ = 32.00 → total per acetate = 59.044
But there are 4 acetates: 4 × 59.044 = 236.176
Plus Pb: 207.2
Total = 207.2 + 236.176 = 443.376 g/mol

Mass = 6.7 × 10⁻² g = 0.067 g

Moles = 0.067 g / 443.376 g/mol ≈ ?

0.067 ÷ 443.376 ≈ 0.0001511 mol

Volume = 3.5 mL = 0.0035 L

M = 0.0001511 mol / 0.0035 L ≈ 0.04317 M

Sig figs: 6.7 × 10⁻² has 2 sig figs, 3.5 mL has 2 sig figs → answer should have 2 sig figs.

→ 0.043 M

✔ Check: 0.067 / 443.376 ≈ 0.0001511; / 0.0035 = 0.04317 → rounds to 0.043 M

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Problem 6:

> I have two solutions. In the first solution, 1.0 moles of sodium chloride is dissolved to make 1.0 liters of solution. In the second one, 1.0 moles of sodium chloride is added to 1.0 liters of water. Is the molarity of each solution the same? Explain your answer.

Important distinction!

- First solution: “dissolved to make 1.0 liters of solution” → final volume is 1.0 L → M = 1.0 mol / 1.0 L = 1.0 M

- Second solution: “added to 1.0 liters of water” → this means you start with 1.0 L water, then add solute. When you add solid to liquid, the total volume increases slightly. So final volume > 1.0 L → molarity < 1.0 M

Therefore, no, they are not the same.

The first is exactly 1.0 M. The second is less than 1.0 M because the total volume is more than 1.0 L.

Example: If you add 1 mole of salt to 1 L water, the volume might become about 1.02 L or so → M ≈ 1.0 / 1.02 ≈ 0.98 M

✔ Answer: No, because in the second case, the total volume is greater than 1.0 L, so concentration is lower.

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## Final Answers:

1) 10 M
2) 0.17 M
3) 200 M
4) 2.67 M
5) 0.043 M
6) No, the molarities are not the same. In the first solution, the final volume is 1.0 L, giving 1.0 M. In the second, adding solute to 1.0 L of water makes the total volume larger than 1.0 L, so molarity is less than 1.0 M.