Question

I'm having trouble determining how to find delta T and the freezing point depression for Solutions #1 and #2. This is all the information I have. Please show all work. Thank you.

Mass of test tube	24.7795 g
Mass of tube + BZP	34.2490 g
Mass of BZP used	9.4695 g
Freezing point of pure BZP	-9.80 kg/mL

	Solution 1	Solution 2
Mass of container and Sample	11.7366 g	11.7366 g
Mass of container and Sample (after weighing)	11.0103 g	9.6156 g
Mass of unknown used	0.7263 g	2.1210 g
Freezing Point of solution	41.2 (C)	39.8 (C)
Freezing Point depression (delta T)
Molality of unknown
Molar Mass of unknown
Average molar mass of unknown

Freezing point of BZP is listed above...-9.80 C kg/Mol

the freezing point of pure BZP was 45.6 celsius

Answer 1

freezing point of pure BZP was 45.6 celsius

freezing point of solution was 41.2 celsius

hence depresssion in freezing point deltaT = |41.2 - 45.6| = 4.4 celsius (solution 1)

for solution 2 , 39.8 - 45.6 = 5.8 celsius

deltaT = Kf * m

where for water Kf = 1.853

hence m = 4.4 / 1.853 = 2.37 m (solution 1)

for solution 2 5.8 / 1.853 = 3.13 m