need help again with freezing point.?

Question

How many grams of sulfur (S8) would have to be dissolved in 105.0 g of carbon tetrachloride to lower the freezing point by 1.70 degrees celsius?

kf = 29.8 K-kg/mol for CCl4

i cant find an example in the text

Answer 1

ΔTf= Kf*m

m= molality =mole solute/kg solvent =
=1000*mole solute / mass(solvent) .. .. with mass solvent expressed in grams

but mole solute =mass solute/ MW solute

so m= 1000*mass(solute)/ (MW*mass(solvent))

when both mass solute and mass solvent are expressed in grams. If we expresses mass solute in grams and mass solvent in kg then we wouldn't multiply by 1000.

So

ΔTf= Kf*mass(solute) *1000 /(MW*mass (solvent)) =>

mass(solute) =MW*mass(solvent) *ΔTf /(1000*Kf)

Since you consider the formula S8 for sulfur MW=8*32 =256

mass(solute) = 256*105*1.7 / (1000*29.8) =1.533 g

Answer 2

I have that same question except at 1.3 degrees celsius. I can't figure it out so far.

http://capahelp101.proboards33.com/index.cgi?board=che132

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