Ask Question
5 July, 20:44

What is the vapor pressure (in mm Hg) at 30∘C of a solution prepared by dissolving 30.5 g of acetone in 23.5 g of ethyl acetate?

+4
Answers (1)
  1. 5 July, 22:47
    0
    The vapor pressure of the solution is 284.29 mmHg

    Explanation:

    From Raoult's law

    Vapor pressure of solution = mole fraction of solvent * vapor pressure of solvent

    Mass of solute (acetone) = 30.5 g

    MW of acetone (CH3COCH3) = 58 g/mol

    Number of moles of solute = mass/MW = 30.5/58 = 0.526 mol

    Mass of solvent (ethyl acetate) = 23.5g

    MW of ethyl acetate (CH3COOC2H5) = 88 g/mol

    Number of moles of solvent = mass/MW = 23.5/88 = 0.267 mol

    Volume of solvent = 0.267*22.4*1000 = 5980.8 cm^3

    Vapor pressure of solvent (P) = nRT/V

    n is number of moles of solvent = 0.267 mol

    R is gas constant = 82.057 cm^3. atm/mol. K

    T is temperature of solution = 30°C = 30+273 = 303 K

    V is volume of the solvent = 5980.8 cm^3

    P = 0.267*82.057*303/5980.8 = 1.11 atm = 1.11*760 = 843.6 mmHg

    Total moles of solution = moles of solute + moles of solvent = 0.526 + 0.267 = 0.793 mol

    Mole fraction of solvent = moles of solvent/moles of solution = 0.267/0.793 = 0.337

    Vapor pressure of solution = 0.337 * 843.6 mmHg = 284.29 mmHg
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “What is the vapor pressure (in mm Hg) at 30∘C of a solution prepared by dissolving 30.5 g of acetone in 23.5 g of ethyl acetate? ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers