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0.845g of the unknown dissolved in 4.250 grams of camphor had a freezing point of 1.36C. The freezing point depression constant for camphor is 40C*kg/Mol and the melting point of pure camphor is 129C.

Calculate the freezing temperature depression, T? Tf=Kf*m

Calulate the molality? Which i know is mol/kg solvent

Calculate the number of moles present of the unknown compound?

Calculate the molar mass of the unknown compound?

Thank you in advance and please lists steps if possible. I really want to understand and to get this!! :)

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marked as duplicate by Klaus-Dieter Warzecha, Freddy, jonsca♦ Sep 20 '14 at 6:52

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    $\begingroup$ $T_f$ is really $\Delta T_f$, and you have K. See if you can get a bit further and then someone can help steer you in the right direction. Remember you're solving for g/mol $\endgroup$ – jonsca♦ Sep 20 '14 at 2:24
  • $\begingroup$ Okay here is what I got Part 1: 1.36-40 is 38.64kg/mol= 0.966 Then for molality .875moles/0.00425kg =205.9 hg/mol Then the number of moles present of unknown compound i got: 0.845g/ 0.966= .875 Then for the fianal one i got for molar mass was 0.845g/.875mol= 1g/mol. please let me know if i am correct $\endgroup$ – Mitchel Johnson Sep 22 '14 at 0:24
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Mass of solute $W_b=0.845\times10^{-3}\;kgm$
Mass of solvent $W_a=4.250\times10^{-3}\;kgm$
Depression in freezing point $\Delta T_f=1.36\;K$
Depression constant for solvent $K_f=40\;K.kg.mol^{-1}$
Freezing point of Camphor $T_f=129\;^\circ C$

Part I: $\Delta T_f$

Part II: Molality($m$): $$\Delta T_f=K_fm\implies m=\frac{\Delta T_f}{K_f}$$

Part III: Moles($n_b$): $$m=\frac{n_b}{W_a}\implies n_b=W_am=W_a\frac{\Delta T_f}{K_f}$$

Part IV: Molecular Mass of solvent($M_b$): $$n_b=\frac{W_b}{M_b}\implies M_b=\frac{W_b}{n_b}=\frac{W_bK_f}{W_a\Delta T_f}$$

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  • $\begingroup$ Okay here is what I got Part 1: 1.36-40 is 38.64kg/mol= 0.966 Then for molality .875moles/0.00425kg =205.9 hg/mol Then the number of moles present of unknown compound i got: 0.845g/ 0.966= .875 Then for the fianal one i got for molar mass was 0.845g/.875mol= 1g/mol. please let me know if i am correct – $\endgroup$ – Mitchel Johnson Sep 22 '14 at 14:14

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