I am only concerned with Part b. I need to have a fluid understanding of the met
ID: 896560 • Letter: I
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I am only concerned with Part b. I need to have a fluid understanding of the method used to solve this problem. Thus, please do not abbreviate your answers. I would prefer any answer to be completely written out on paper or typed in Latex. Also, I have the answer, but I am having a difficulty time following how it was determined. Thanks in Advance.
6 1. a) (5 pts)) The attractive forces between molecules of a substance and their The temperature where all three phases of point. Raoult's law of the container is called a substance are is equilibrium is called the involves a relationship between the mole fraction and the gas in a solution. The abbreviation for the concentration unit of the ratio of solute mass to solution mass times one billion is increases most gases become . When the temperature soluble in solution. b) 6 pts) Methanol, CH3OH, has specific heat of 0.328 cal/g C for its vapor and its heat of vaporization is 35.21 kJ/mole at its boiling point of 64.60 °C. Calculate the specific heat in J/g·°C of liquid methanol if 122.1 kJ of heat is required to convert 100.0 g of liquid methanol at 32.60 °C to vapor at 95.60 °C. At wts: C-12.00 g/mol, H= 1.008 g/mol and O= 16.00 g/mol. 1 cal=4.1840 J.Explanation / Answer
b) We do it it parts,
total heat q required to convert 100 g of liq. MeOH at 32.60 oC to vapor at 95.60 oC = 122.1 kJ = 122.1 x 1000 J
Convert liquid MeOH at 32.60 oC to liquid MeOH at 64.60 oC
q1 = mCpdT = 100 x Cp x (64.6-32.6)
Convert liquid MeOH to vapor at 64.60 oC
q2 = mdHvap = 100 x 35.21 x 1000/32.04 = 1.10 x 10^5 J
Convert vapor MeOH at 64.60 oC to vapor at 95.60 oC
q3 = mCpdT = 100 x 0.328 x 4.1840 x (95.60-64.60) = 4254.2912 J
Add q1, q2 and q3, the total showuld be q
122.1 x 1000 = 100 x Cp x (64.6-32.6) + 1.10 x 10^5 + 4254.2912
Cp = 2.53 J/g.oC
So the specific heat for liquid MeOH would be 2.53 J/g.oC
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