Example: Calculate AP, 4Tf, Tb and TT for solutions which are 1% by weight in Benzene of solutes with M=10² and 105 Data: at 25°c, P₁° = 95.9 Torr 4H4 = 10.6 kJ/mole 4Hy = 30.8 kJ/mole (= 0.879 9/m³ Results: M₂/n, M₂ = 10² 7.88×103 T° = 5.5°C Tb° = 80.1°C M₂=105 1.88x106 c: weight concentration. AT lim 2-0 C ΔΤ, lim 2-0 C RT2 PAH, M G.) lim- c-0 C RT2 ΡΔΗ, M₁ I RT M₁ Pi-Pi Pi -=X₂

Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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this is Polymer properties problem
Example:
Calculate AP, 4Tf, Tb and π for solutions which are
1% by weight in Benzene of solutes with M=10² and 105
Data: at 25°c, P₁° = 95.9 Torr
4H4 = 10.6 kJ/mole
4Hy = 30.8 kJ/mole
(= 0.879 9/m³
Results:
M₂/n,
M₂ = 10²
7.88×103
T° = 5.5°C
Tb° = 80.1°C
M₂=105
188x106
c: weight concentration.
AT
lim
2-0 C
AT
lim
2-0 C
RT2
PAH, M
I
lim-
c-0 C
G.)
RT2
ΡΔΗ,
RT
M₁
Pi-Pi
Pi
= X₂
M₁
Transcribed Image Text:Example: Calculate AP, 4Tf, Tb and π for solutions which are 1% by weight in Benzene of solutes with M=10² and 105 Data: at 25°c, P₁° = 95.9 Torr 4H4 = 10.6 kJ/mole 4Hy = 30.8 kJ/mole (= 0.879 9/m³ Results: M₂/n, M₂ = 10² 7.88×103 T° = 5.5°C Tb° = 80.1°C M₂=105 188x106 c: weight concentration. AT lim 2-0 C AT lim 2-0 C RT2 PAH, M I lim- c-0 C G.) RT2 ΡΔΗ, RT M₁ Pi-Pi Pi = X₂ M₁
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