Explore BrainMass
Share

Explore BrainMass

    Molality, boiling and freezing of solutions: Example questions

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    What is the freezing point, boiling point, and molality of a solution with 150 grams of aluminum chloride dissolved in 350 mL of water?

    How much ethyl alcohol, C_2H_5OH is needed to dissolved 75 grams of naphthalene, C_10H_8 and produce a 0.1 molal solution?

    How many litres of water would be needed to dissolve 25 grams of sodium hydroxide and make a 0.6 molal solution, and what would be the freezing and boiling point of the solution?

    How much benzene would be needed to dissolve 250 grams of bromine and produce a 1.4 molal solution?

    How many kilograms of water would be needed to dissolve 1000 grams of aluminum sulfate and produce a 7.3 molal solution and what would be the freezing and boiling point of the solution?

    How many grams, millilitres, kilograms, and litres of water would be needed to dissolve 85 grams of potassium nitrate and produce a 0.5 molal solution and what would be the freezing and boiling point of the solution?

    © BrainMass Inc. brainmass.com October 10, 2019, 2:51 am ad1c9bdddf
    https://brainmass.com/chemistry/stoichiometry/molality-boiling-freezing-solutions-example-questions-396864

    Solution Preview

    MOLALITY, BOILING AND FREEZING OF SOLUTIONS:

    Important Notes to Remember:
    *Molality = m = mole of solute / kg of solvent
    *Solutions have lower freezing point and higher boiling point as compared to the pure solvent.
    *To compute for the freezing point depression, use the equation:
    Tf = Kfm(i)
    Where :
    Tf = freezing point depression
    Kf = freezing point depression constant (specific for a given solvent)
    (For H2O: Kf = 1.86 0C / m)
    m = molality of solution
    i = vant Hoff factor
    If the solute is nonelectrolyte, i = 1
    If the solute is an electrolyte, i = number of ions produced per mole of salt
    dissolved in solution
    *To compute for the freezing point of the solution, use the equation:
    Freezing point of the solution = freezing point of the solvent - freezing point
    depression
    *To compute for the boiling point elevation, use the equation:
    Tb = Kbm(i)
    Where:
    Tb = boiling point elevation
    Kb = boiling point elevation constant (specific for a given solvent)
    (For H2O: Kb = 0.512 0C / m)
    m = molality of solution
    i = vant Hoff factor
    If the solute is nonelectrolyte, i = 1
    If the solute is an electrolyte, i = number of ions produced per mole of salt
    dissolved in solution
    *To compute for the boiling point of the solution, use the equation:
    Boiling point of the solution = boiling point of the solvent + boiling point elevation

    1. What is the freezing , boiling point and molality of a solution with 150 grams of aluminum chloride dissolved in 350 ml of water?

    Molality = m = mole solute / kg solvent

    Mole AlCl3 = 150 g (1 mole AlCl3 / 133.34 g)
    Mole AlCl3 = 1.125

    Solving for kilogram of water:
    (Note: ...

    Solution Summary

    The solution answers questions posted regarding calculating the freezing point, boiling point and molality.

    $2.19