How many grams of sulfur (S8) would have to be dissolved in 120.0 g of carbon tetrachloride to lower the freez

Question

How many grams of sulfur (S8) would have to be dissolved in 120.0 g of carbon tetrachloride to lower the freezing point by 1.70 degrees celsius?


kf = 29.8 K-kg/mol for CCl4

Answer 1

I'll set up, you can crunch numbers.
Find moles to depress freeze point of 1 kg CCL4
by 1.7 C: 1.7 /29.8 = 0.06(about)
Find g of S8 corresponding to that number of moles (0.06 x [32x8])= 17 (about)
Now scale that to the amount of CCl4:
17 x (120/1000) = about 2

Answer 2

Tf = Kf*Cm
1.70 = 29.8( x moles S8/ .120kg CCl4)
x moles S8 = 6.85x10^-3 mol
6.85x10^-3 mol S8 * (256.56 g S8 / 1 mol S8) = 1.76g S8

1.76g S8 must be dissolved in 120.0g of CCl4 in order to lower the freezing point by 1.70C.