Practice problems on calculating concentrations of solutions in chemistry.

Let's solve each of these practice problems step by step. I'll go through them one at a time and explain the concepts and calculations involved.

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1. What is the mass percent of a solution of 7.6 grams sucrose in 83.4 grams of water?

Mass percent = (mass of solute / mass of solution) × 100
- Mass of solute (sucrose) = 7.6 g
- Mass of solvent (water) = 83.4 g
- Mass of solution = 7.6 + 83.4 = 91.0 g

$$
\text{Mass percent} = \frac{7.6}{91.0} \times 100 = 8.35\%
$$

✔ Answer: 8.35%

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2. How many grams of sucrose must be added to 375 grams of water to prepare a 2.75 % by mass solution of sucrose?

Let $ x $ = mass of sucrose needed
Total mass of solution = $ x + 375 $ g

$$
\frac{x}{x + 375} = \frac{2.75}{100}
$$

Multiply both sides:

$$
100x = 2.75(x + 375)
$$
$$
100x = 2.75x + 1031.25
$$
$$
97.25x = 1031.25
$$
$$
x = \frac{1031.25}{97.25} \approx 10.60 \, \text{g}
$$

✔ Answer: 10.6 grams

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3. A saline solution, NaCl in water, is 0.92 % (m/v). How many grams of NaCl are required to prepare 35.0 mL of this solution?

(m/v)% = (mass of solute / volume of solution) × 100

So:
$$
\frac{\text{mass of NaCl}}{35.0} = \frac{0.92}{100}
\Rightarrow \text{mass of NaCl} = \frac{0.92 \times 35.0}{100} = 0.322 \, \text{g}
$$

✔ Answer: 0.322 grams

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4. What is the molarity of 4.35 moles KMnO₄ dissolved in 750 mL of solution?

Molarity (M) = moles / liters

Volume = 750 mL = 0.750 L

$$
M = \frac{4.35}{0.750} = 5.80 \, \text{M}
$$

✔ Answer: 5.80 M

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5. What is the molarity of 20.0 grams of NaOH dissolved in 1.50 L of solution?

First, find moles of NaOH:
Molar mass of NaOH = 23 + 16 + 1 = 40.0 g/mol

$$
\text{moles} = \frac{20.0}{40.0} = 0.500 \, \text{mol}
$$

$$
M = \frac{0.500}{1.50} = 0.333 \, \text{M}
$$

✔ Answer: 0.333 M

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6. How many grams of KNO₃ are present in 185 mL of a 2.50 M solution?

Convert mL to L: 185 mL = 0.185 L

Moles = M × V = 2.50 × 0.185 = 0.4625 mol

Molar mass of KNO₃ = 39 + 14 + 48 = 101 g/mol

Mass = 0.4625 × 101 = 46.7125 g ≈ 46.7 g

✔ Answer: 46.7 grams

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7. How many mL of a 0.10 M FeSO₄ solution are required to provide 0.35 g of FeSO₄?

Molar mass of FeSO₄ = 56 + 32 + 64 = 152 g/mol

Moles = $ \frac{0.35}{152} = 0.002303 $ mol

Volume (L) = moles / M = $ \frac{0.002303}{0.10} = 0.02303 $ L = 23.0 mL

✔ Answer: 23.0 mL

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8. How many mL of a 0.300 M AgNO₃ solution will it take to make 500 mL of a 0.100 M AgNO₃ solution?

Use dilution formula: $ M_1V_1 = M_2V_2 $

$ M_1 = 0.300 $, $ V_1 = ? $, $ M_2 = 0.100 $, $ V_2 = 500 $ mL

$$
V_1 = \frac{0.100 \times 500}{0.300} = \frac{50}{0.3} = 166.7 \, \text{mL}
$$

✔ Answer: 167 mL (rounded)

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9. A solution contains 128 g of CH₃OH and 108 g of water. What is the mole fraction of CH₃OH?

Molar mass CH₃OH = 12 + 4(1) + 16 = 32 g/mol
Molar mass H₂O = 18 g/mol

Moles CH₃OH = $ \frac{128}{32} = 4.0 $ mol
Moles H₂O = $ \frac{108}{18} = 6.0 $ mol
Total moles = 4.0 + 6.0 = 10.0 mol

Mole fraction of CH₃OH = $ \frac{4.0}{10.0} = 0.400 $

✔ Answer: 0.400

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10. What mass of glucose, C₆H₁₂O₆, must be dissolved in 150.0 g of water so that the mole fraction of glucose is 0.125?

Let $ x $ = mass of glucose
Molar mass glucose = 180 g/mol
Moles glucose = $ \frac{x}{180} $

Moles water = $ \frac{150.0}{18} = 8.333 $ mol

Mole fraction of glucose = $ \frac{\text{moles glucose}}{\text{total moles}} = 0.125 $

$$
\frac{x/180}{x/180 + 8.333} = 0.125
$$

Let $ y = x/180 $, then:

$$
\frac{y}{y + 8.333} = 0.125
\Rightarrow y = 0.125(y + 8.333)
\Rightarrow y = 0.125y + 1.0416
\Rightarrow 0.875y = 1.0416
\Rightarrow y = 1.190
$$

Then $ x = 1.190 \times 180 = 214.2 $ g

✔ Answer: 214 g

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11. What mass of water must be used to prepare a solution of 25.5 g of CaCl₂ dissolved in water if the mole fraction of CaCl₂ is 0.105?

Molar mass CaCl₂ = 40 + 71 = 111 g/mol
Moles CaCl₂ = $ \frac{25.5}{111} = 0.2297 $ mol

Let $ w $ = mass of water in grams
Moles water = $ \frac{w}{18} $

Mole fraction of CaCl₂ = $ \frac{0.2297}{0.2297 + w/18} = 0.105 $

Solve:

$$
\frac{0.2297}{0.2297 + w/18} = 0.105
\Rightarrow 0.2297 = 0.105(0.2297 + w/18)
\Rightarrow 0.2297 = 0.02412 + 0.105(w/18)
\Rightarrow 0.20558 = 0.105(w/18)
\Rightarrow w/18 = \frac{0.20558}{0.105} = 1.958
\Rightarrow w = 1.958 \times 18 = 35.24 \, \text{g}
$$

✔ Answer: 35.2 g

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12. What is the molality of a solution that contains 46 g of CH₃OH dissolved in 348 g of water?

Molality (m) = moles of solute / kg of solvent

Moles CH₃OH = $ \frac{46}{32} = 1.4375 $ mol
Mass water = 348 g = 0.348 kg

$$
m = \frac{1.4375}{0.348} = 4.13 \, \text{m}
$$

✔ Answer: 4.13 m

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13. What mass of AgNO₃ must be dissolved in 200 g of water to prepare a 0.250 m solution?

Molality = moles / kg solvent → moles = molality × kg solvent

$ \text{moles} = 0.250 \times 0.200 = 0.0500 $ mol

Molar mass AgNO₃ = 108 + 14 + 48 = 170 g/mol

Mass = $ 0.0500 \times 170 = 8.50 $ g

✔ Answer: 8.50 g

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14. If an aqueous solution of urea, NH₂CONH₂, is 26.0 % by mass and has a density of 1.07 g/mL, calculate the molality of urea in solution.

Assume 100 g solution:
- Mass urea = 26.0 g
- Mass water = 74.0 g = 0.0740 kg

Molar mass urea = 60 g/mol
Moles urea = $ \frac{26.0}{60} = 0.4333 $ mol

Molality = $ \frac{0.4333}{0.0740} = 5.85 \, \text{m} $

✔ Answer: 5.85 m

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15. What is the percent by mass of methanol, CH₃OH, if the mole fraction of methanol dissolved in water is 0.500?

Let mole fraction of CH₃OH = 0.500 ⇒ moles CH₃OH = moles H₂O = 1 mol (assume)

Molar mass CH₃OH = 32 g/mol → mass = 32 g
Molar mass H₂O = 18 g/mol → mass = 18 g
Total mass = 32 + 18 = 50 g

Mass percent = $ \frac{32}{50} \times 100 = 64.0\% $

✔ Answer: 64.0%

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16. Calculate the molarity of a solution that is 39.77 % H₂SO₄ by mass. The density of the solution is 1.305 g/mL.

Assume 1 L (1000 mL) of solution:

Mass of solution = $ 1.305 \times 1000 = 1305 $ g
Mass of H₂SO₄ = $ 0.3977 \times 1305 = 518.8 $ g

Molar mass H₂SO₄ = 98 g/mol
Moles = $ \frac{518.8}{98} = 5.294 $ mol

Molarity = $ \frac{5.294}{1} = 5.29 \, \text{M} $

✔ Answer: 5.29 M

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17. What is the molality of a solution that contains 128 grams of CH₃OH in 108 grams of water?

Moles CH₃OH = $ \frac{128}{32} = 4.0 $ mol
Mass water = 108 g = 0.108 kg

Molality = $ \frac{4.0}{0.108} = 37.0 \, \text{m} $

✔ Answer: 37.0 m

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18. Calculate the weight percent of HCl in 3.20 M solution. The density of the solution is 1.10 g/mL.

Assume 1 L = 1000 mL

Mass of solution = $ 1.10 \times 1000 = 1100 $ g
Moles HCl = 3.20 mol
Molar mass HCl = 36.5 g/mol
Mass HCl = $ 3.20 \times 36.5 = 116.8 $ g

Weight percent = $ \frac{116.8}{1100} \times 100 = 10.62\% $

✔ Answer: 10.6%

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19. Calculate the molality of Ca(OH)₂ in a 1.50 M aqueous solution that has a density of 1.320 g/mL.

Assume 1 L = 1000 mL

Mass of solution = $ 1.320 \times 1000 = 1320 $ g
Moles Ca(OH)₂ = 1.50 mol
Molar mass Ca(OH)₂ = 40 + 32 + 2 = 74 g/mol
Mass Ca(OH)₂ = $ 1.50 \times 74 = 111 $ g

Mass of water = 1320 - 111 = 1209 g = 1.209 kg

Molality = $ \frac{1.50}{1.209} = 1.24 \, \text{m} $

✔ Answer: 1.24 m

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20. What is the mole fraction of commercial "concentrated" hydrofluoric acid, which is 55.0 % HF by mass?

Assume 100 g solution:
- Mass HF = 55.0 g
- Mass H₂O = 45.0 g

Molar mass HF = 20 g/mol → moles = $ \frac{55.0}{20} = 2.75 $ mol
Molar mass H₂O = 18 g/mol → moles = $ \frac{45.0}{18} = 2.50 $ mol
Total moles = 2.75 + 2.50 = 5.25 mol

Mole fraction HF = $ \frac{2.75}{5.25} = 0.524 $

✔ Answer: 0.524

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## ✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | 8.35% |
| 2 | 10.6 g |
| 3 | 0.322 g |
| 4 | 5.80 M |
| 5 | 0.333 M |
| 6 | 46.7 g |
| 7 | 23.0 mL |
| 8 | 167 mL |
| 9 | 0.400 |
| 10 | 214 g |
| 11 | 35.2 g |
| 12 | 4.13 m |
| 13 | 8.50 g |
| 14 | 5.85 m |
| 15 | 64.0% |
| 16 | 5.29 M |
| 17 | 37.0 m |
| 18 | 10.6% |
| 19 | 1.24 m |
| 20 | 0.524 |

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