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Chemistry

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The freezing point depression of a 0.100 molality solution of NaCl(aq) solution is 0.34 deg C. Calculate the % dissociation of NaCl(aq) (Kf for water = 1.86 deg Cm^-1)

  • Chemistry - ,

    Do you mean -0.34 degrees C. I don't see how it could be +0.34.

  • Chemistry - ,

    yeah -.34 is the change in temp freezing

  • Chemistry - ,

    delta T = i*Kf*m
    0.34 = i*1.86*0.1
    solve for i.
    Then i x 0.1 = effective molality which I will call em.
    NaCl <==>Na^+ + Cl^-=
    Do an ICE chart for NaCl
    NaCl will be 0.1-y
    Na^+ will be y
    Cl^- will be y
    The sum of NaCl, Na^+, and Cl^- will be equal to em
    So you will have
    0.1-y + y + y = em
    solve for y
    Then (y/i)*100 = % ionization.
    Post your work if you get stuck.

  • Chemistry - ,

    i got an i of 1.83 then i divided it by 2 to get a dissociation of 91.5% is this correct?

  • Chemistry - ,

    The 1.83 for i is correct. But my instructions didn't say anything about dividing by 2. Follow my instructions calculate effective molality, then go through the ICE chart, then to y, then to percent ionization.

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