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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: zerofantasym on February 05, 2008, 12:56:02 PM

Title: Freezing point depression problem
Post by: zerofantasym on February 05, 2008, 12:56:02 PM

A solution was prepared by mixing 2.17 g of unknown non-electrolyte with 225.0 g of chloroform. The freezing point of the resulting solution is -64.2 C. The freezing point of pure chloroform is -63.5 C, and it's k(f) = 4.68 C. What is the molecular mass of the unknown.

 I keep getting 64.2 g/mol but the answer in the book is  69.3 g/mol
∆°
∆

∆Tf= -63.5° - (-64.2°) = 0.70 °

m= .70°/4.68 ° (m^-1) = 0.150 m

mol solute = .150 m x (225.0 x 10^-3) kg = 33.8 x 10^-3 mol

molecular weight = 2.17 g X / (33.8 x 10^-3) = 64.2 g/mol


What is going on??? :( Help :(

Title: Re: Freezing point depression problem
Post by: enahs on February 05, 2008, 02:30:36 PM

Without any rounding errors, I get 64.48 g/mol.

Are you getting the K_f and Freezing point of pure Chloroform from your book or the internet?

Title: Re: Freezing point depression problem
Post by: zerofantasym on February 05, 2008, 10:22:02 PM

book

Title: Re: Freezing point depression problem
Post by: enahs on February 05, 2008, 10:35:58 PM

Books are sometimes wrong. Check the publishers website for an errata and see if it has correction to that problem.

Title: Re: Freezing point depression problem
Post by: zerofantasym on February 09, 2008, 10:57:50 AM

hmm turns out the book was wrong.