Chemistry help needed badly

Question

1. Carbon tetrachloride (CCL4, MW 153.81, Kb= 5.03 degrees C/m and BP=77.2 degrees C, d= 1.59 g/mL) is an excellent solvent for chlorinated compounds. When 5.00 g of trichloromethane, CHCl3 (MW 119.37) is dissolved in 100. g CCl4, at what temperature will the solution boil?

2. A saline solution in an IV bag is listed as 0.90% m/v NaCl (ME 58.45 g/mol.) The density of the solution is 1.186 g/mL. Determine the Molarity and the molality of the solution.

Thanks!

Answer 1

∆T = imK

∆T = change in temperature or change in boiling point

i = van't Hoff factor = 1 for CHCl3

m = molality = moles CHCl3/kg CCl4 

K = boiling point constant for CCl4 = 5.03º/m

First, calculate m and then substitute in above equation:

moles CHCl3 = 5.00 g x 1 mole/119.37 g = 0.04189 moles

kg CCl4 = 100 g CCl4 = 0.1 kg

molality = 0.04189 moles/0.1 kg = 0.4189 m

∆T = (1)(0.4189 m)(5.03º/m) = 0.2107 degrees C

New boiling point = 0.2107 + 77.2 = 77.4 degrees C

NOTE:  density of CCl4 is not needed for this problem.

 

0.90% m/v = 0.90 g/100 ml soln = 9.0 g/liter soln

9.0 g/L x 1 mole/58.45 g = 0.15398 moles/L = 0.154 M = molarity

molality = moles NaCl/kg water

1.186 g/ml = 1186 g/L

1186 g - 9 g = 1177 g H2O = 1.177 kg H2O

m = 0.154 moles NaCl/1.177 kg H2O = 0.1308 m = 0.131 m 

Answers given to 2 significant figures (based on 0.90%) would be 0.15 M and 0.13 m.