Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Freezing point

  1. Nov 28, 2004 #1
    How many grams of urea (MW = 60.056 g/mol) would have to be dissolved in 115.0 grams of water to lower the freezing point by 1.60 degrees celsius?

    Equation for freezing point
    Tf= Kf x Cm

    Kf= 1.858
    Cm= molality of solute in kg

    please help, iam getting the same answer
  2. jcsd
  3. Nov 28, 2004 #2


    User Avatar
    Science Advisor
    Gold Member

    Here, [itex]T_f[/itex] is 1.60, and from there you can calculate the molality:

    [tex]T_f=K_f \times C_m and C_m=\frac {T_f}{K_f}[/tex]

    [tex]C_m=\frac {1.60}{1.858}=0.861[/tex]

    Since your molality is 0.861, it means that in 1 kg of solution you'll have to dissolve moles of this amount of compound.
  4. Nov 28, 2004 #3
    i still dont kn ow how to calculate the grams required, please help
  5. Nov 29, 2004 #4


    User Avatar
    Science Advisor
    Gold Member

    What did you find as mole amount from the last equation I gave? You'll just multiply this value with molar mass of urea to learn how many grams should be in the solution.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Freezing point