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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


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    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


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    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.
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