How many grams of sulfur (S8) would have to be dissolved in 150.0 g of carbon?

Question

How many grams of sulfur (S8) would have to be dissolved in 150.0 g of carbon tetrachloride to lower the freezing point by 1.50 degrees celsius?


kf = 29.8 K-kg/mol for CCl4

Answer 1

Again, please see my previous post that has this exact problem solved.

http://answers.yahoo.com/question/index;_ylt=Al49CHqp9NV_tc18fq3hNFbty6IX?qid=20070307103312AAvIRks&show=7#profile-info-cU6rtQyraa

Answer 2

I guess you can use my question as a ref.

How many grams of sulfur (S8) would have to be dissolved in 110.0 g of carbon tetrachloride to lower the freezing point by 1.70 degrees celsius?

kf = 29.8 K-kg/mol for CCl4

Enter a numerical answer only, do not include units in the answer you enter

Correct, computer gets: 1.60945503355705

Hint: Don't forget to include the van't Hoff factor i, if it is needed.
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This is how I solve it:

(1.70)*(0.110)/(1)(29.8)=.0062751678

(32.066)*(8 atoms)*(.0062751678)=1.609756245

Answer 3

Tf = Kf*Cm a million.70 = 29.8( x moles S8/ .120kg CCl4) x moles S8 = 6.85x10^-3 mol 6.85x10^-3 mol S8 * (256.fifty six g S8 / a million mol S8) = a million.76g S8 a million.76g S8 ought to be dissolved in one hundred twenty.0g of CCl4 so as to decrease the freezing element via a million.70C.