chemistry

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When 0.500 g of an unknown compound was dissolved in 15.0 g benzene, the freezing point depression was determined to be 0.320 C. The molar mass of the unknown compound is ____

(The freezing point depression constant for benzene is 5.12 C kg/mol. )

  • chemistry -

    well, you know the freezing point depression is equal to mass/molmass *k, and you know grams, and the constant k.

  • chemistry -

    would the answer be .305? i'm still getting this wrong.

    i set up the equation to be

    x = .500 g/ (.320 C) (5.12 C)

  • chemistry -

    would the answer be .305? i'm still getting this wrong.

    i set up the equation to be

    x = .500 g/ (.320 C) (5.12 C)

  • chemistry -

    I erred in my molality formula

    5.12C= .5(mmass*kgbenzene) * .320

    mmass= .5*.320/(5.12*.015)

    I get about 50 for the molmass.

  • chemistry -

    ah i got 2.08 and im still getting it wrong

  • chemistry -

    Isn't delta T = Kb*m?
    Then 0.320 = (5.12oC/m)*m
    m = 0.320/5.12 = 0.0625 m

    molality = mols/kg so
    mols = molality*kg = 0.0625 mol/kg*0.015 kg = 9.375 x 10^-4 mols.

    mols = g/molar mass
    molar mass = g/mols = 0.500 g/9.375 x 10^-4 = 533.3

  • chemistry -

    thank you

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