Freezing point depression

Opejay

Full Member
May 30, 2012
18
0
  1. Pre-Medical
    When 10 grams of a certain material are dissolved in 1 kg of water, the freezing point of the solution is -0.46 Celsius. Which of the following could be the substance? ( water has a freezing point depression constant of 1.86K/m)

    A. HC2H3O2

    B. HNO3

    C. NACl

    D. CaCl2

    thanks
     

    gongshow89

    Full Member
    Mar 19, 2011
    28
    1
    1. Pre-Medical
      Looks like it's D..

      We know water normally has a freezing/melting point of 0. So, if it is now -0.46 then we need to determine which substrate would give this.

      The formula to use is deltaT=k*b*i

      deltaT is 0.46 in this example as stated above. i is the number of ion particles, and b is the molality.

      In this case we have 10 g/1kg of solvent (water). If you use CaCl2 then the molar mass is ~110 g/mole which means we have about 0.09 moles/kg of water. (10/110 = 0.09)

      Since there are 3 ion particle in CaCl2 we see that the product of 0.09 * 3 * 1.86 (constant in the formula)= ~0.5 which is close to the 0.46 celsius change. This answer gave me the closest value.

      Hope that matches up with the answer in the book.. I take it this is from EK? There may be a non-mathematical way of doing this, but this is the way I worked it out.
       

      sciencebooks

      Full Member
      10+ Year Member
      Jul 27, 2009
      802
      20
      Michigan
      1. Pre-Medical
        Looks like it's D..

        We know water normally has a freezing/melting point of 0. So, if it is now -0.46 then we need to determine which substrate would give this.

        The formula to use is deltaT=k*b*i

        deltaT is 0.46 in this example as stated above. i is the number of ion particles, and b is the molality.

        In this case we have 10 g/1kg of solvent (water). If you use CaCl2 then the molar mass is ~110 g/mole which means we have about 0.09 moles/kg of water. (10/110 = 0.09)

        Since there are 3 ion particle in CaCl2 we see that the product of 0.09 * 3 * 1.86 (constant in the formula)= ~0.5 which is close to the 0.46 celsius change. This answer gave me the closest value.

        Hope that matches up with the answer in the book.. I take it this is from EK? There may be a non-mathematical way of doing this, but this is the way I worked it out.

        I also just looked at my options (eliminating A immediately as it wouldn't ionize). I wonder if there was a faster way?
         
        This thread is more than 9 years old.

        Your message may be considered spam for the following reasons:

        1. Your new thread title is very short, and likely is unhelpful.
        2. Your reply is very short and likely does not add anything to the thread.
        3. Your reply is very long and likely does not add anything to the thread.
        4. It is very likely that it does not need any further discussion and thus bumping it serves no purpose.
        5. Your message is mostly quotes or spoilers.
        6. Your reply has occurred very quickly after a previous reply and likely does not add anything to the thread.
        7. This thread is locked.