Molarity Practice Worksheet with three problems involving sodium chloride solutions.

Explanation:
Let’s solve each part step by step. Remember:
Molarity (M) = moles of solute ÷ liters of solution

We’ll need the molar mass of sodium chloride (NaCl):
- Na = 23 g/mol
- Cl = 35.5 g/mol
→ NaCl = 23 + 35.5 = 58.5 g/mol

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Problem 1:
0.5 moles of NaCl dissolved to make 0.05 L of solution.

Molarity = moles / liters = 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 NaCl dissolved to make 0.05 L of solution.

First, convert grams → moles:
moles = mass / molar mass = 0.5 g / 58.5 g/mol ≈ 0.008547 mol

Now calculate molarity:
M = 0.008547 mol / 0.05 L ≈ 0.1709 M

Rounded to 3 significant figures (since 0.5 g has 1 sig fig, but often in these problems we keep 2–3 for clarity), we can say ≈ 0.171 M
But let’s keep more precision for now and round at the end if needed.

Double-check:
0.5 / 58.5 = 0.0085470085…
Divide by 0.05 → ×20 → 0.17094… → yes, ~0.171 M

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Problem 3:
0.5 grams of NaCl dissolved to make 0.05 mL of solution.

⚠️ Important: volume is in milliliters, not liters!
Convert mL → L:
0.05 mL = 0.05 / 1000 = 0.00005 L = 5 × 10⁻⁵ L

Moles of NaCl = 0.5 g / 58.5 g/mol ≈ 0.008547 mol (same as before)

Molarity = 0.008547 mol / 0.00005 L = 0.008547 / 5×10⁻⁵
= 0.008547 × (1 / 5×10⁻⁵) = 0.008547 × 20,000 = 170.94 M

That’s extremely high — unrealistic for real solutions (NaCl solubility is ~6 M at room temp), but mathematically correct *if* the problem assumes it dissolves fully.

So:
≈ 171 M (rounded to 3 sig figs)

But again, check calculation:
0.5 / 58.5 = 0.008547
0.008547 ÷ 0.00005 = 170.94 → yes.

Final answers:
1) 10 M
2) 0.171 M
3) 171 M

Let me verify units once more:
- #1: mol/L → correct
- #2: g → mol using 58.5 → then / L → correct
- #3: g → mol, mL → L (÷1000), then / L → correct

All good.

Final Answer:
1) 10 M
2) 0.171 M
3) 171 M