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{solute}}{m{solvent}} where m is in m (molality), n{solute} is moles of solute, and m{solvent} is mass of solvent in kilograms.

  • Mass percent (w/w) definition: \%m = \frac{m{solute}}{m{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{H2O_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{H2O_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{solute} = \frac{m{solute}}{M{H2O_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{solute}}{m{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).