CHE - Ch. 14 Practice Problem

Overview: Hydrogen peroxide in context

• Hydrogen peroxide (H₂O₂) is a powerful oxidizing agent.

• Uses include concentrated solutions in rocket fuels and dilute solutions as hair bleach.

• Example problem: An aqueous solution of H₂O₂ is 30.0% by mass. Molar mass of H₂O₂ is 34.0 g/mol. Calculate its molality.

Key concepts and formulas

• Molality definition: $m = \frac{n<em>{solute}}{m</em>{solvent}}$ where $m$ is in m (molality), $n<em>{solute}$ is moles of solute, and $m</em>{solvent}$ is mass of solvent in kilograms.

• Mass percent (w/w) definition: $\%m = \frac{m<em>{solute}}{m</em>{solution}} \times 100\%$

• To convert from mass percent to masses in a 100 g sample: if solution is 30.0% by mass, then in 100.0 g of solution:

  • Mass of solute (H₂O₂): $m_{solute} = 30.0\ \text{g}$

  • Mass of solvent: $m_{solvent} = 100.0\ \text{g} - 30.0\ \text{g} = 70.0\ \text{g}$

• Molar mass of H₂O₂: $M<em>{H</em>2O_2} = 34.0\ \text{g mol}^{-1}$

• Unit conversions: $1\ \text{kg} = 1000\ \text{g}$

Worked example: 30.0% (w/w) H₂O₂ solution

• Given: $\%m = 30.0\%$, $M<em>{H</em>2O_2} = 34.0\ \text{g mol}^{-1}$

• Assume a 100.0 g sample of solution:

  • Solute mass: $m_{solute} = 30.0\ \text{g}$

  • Solvent mass: $m_{solvent} = 70.0\ \text{g}$

• Compute moles of solute: $n<em>{solute} = \frac{m</em>{solute}}{M<em>{H</em>2O_2}} = \frac{30.0\ \text{g}}{34.0\ \text{g mol}^{-1}} = 0.882\ \text{mol}$

• Convert solvent mass to kilograms: $m_{solvent} = \frac{70.0\ \text{g}}{1000} = 0.0700\ \text{kg}$

• Calculate molality: $m = \frac{n<em>{solute}}{m</em>{solvent}} = \frac{0.882\ \text{mol}}{0.0700\ \text{kg}} = 12.6\ \text{m}$

• Significant figures: 30.0% (3 sig figs) yields molality to 3 sig figs: $m = 12.6\ \text{m}$

Final answer and selection

• Molality: $m = 12.6\ \text{m}$

• Correct option: B

Notes: concepts to remember

• Molality is defined with respect to the mass of solvent, not the solution, and is independent of temperature for its definition (unlike molarity).

• Mass percent conversions rely on assuming a specific sample size (commonly 100 g) to simplify calculations.

• The calculation chain demonstrates converting mass percent to moles of solute and mass of solvent to kilograms before applying the molality formula.

Quick practice tips

• When given mass percent, start by selecting 100 g of solution to make the math straightforward.

• Always convert solvent mass to kilograms before plugging into the molality formula.

• Check units at each step: moles cancel, leaving molality in mol kg⁻¹.

• Use correct significant figures based on the given data (here, 3 sig figs).