2 Chem problems please help if you can?
2) The freezing point for t-butanol is 25.50 degrees Celsius and Kr is 9.1 degrees Celsius/m. Usually t-butanol absorbs water on exosure to air. If the Freezing point of a 10.0g sample of t-butanol is measured as 24.59 degrees Celsius, how many grams of water is present in the sample?
Answer:
1) Pa = P°a Xa and Pb = P°b Xb
Xa = 1.25 / 1.25 + 5.30 = 0.191
Pa = 205 x 0.191 = 39.1 torr
Xb = 5.30 / 1.25 + 5.30 = 0.809
Pb = 135 x 0.809 = 109.2 torr
2) delta T = 25.50 - 24.59 = 0.91
delta T = k x m
0.91 = 9.1 x m
m = 0.1
molar mass t-butanol = 74 g/mol
10.0 / 74 = 0.135 moles
m = 0.1 = 0.135 moles / Kg water
Kg water = 1.35 = 1350 g
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