If somesone could explain these 3 questions and show steps as to how you got the
ID: 1012886 • Letter: I
Question
If somesone could explain these 3 questions and show steps as to how you got the answer, would be greatly appreciated!!
A solution is prepared by dissolving 40.89-grams of ethylene glycol (C2H6O2) in 0.677-kilograms of water. The final volume of the solution is 516-milliliters. For this solution, calculate the concentration as a mole percent (%). The molar mass of ethylene glycol is 62.07 g*mol-1 and the molar mass of water is 18.02 g*mol-1. Don't forget to include the unit symbol in the answer.
What is the vapor pressure (in torr) of 431-mL water, if the heat of vaporation is -10.8-kJ, -constant is 30.5 fraction numerator m L over denominator m o l end fraction , R-constant is 0.00831 fraction numerator k J over denominator m o l cross times K end fraction , and the temperature is 568.13-Kelvin? Don't forget to include your units. Assume this is an Ideal Gas.
Benzene has a heat of vaporization of 30.72 open parentheses fraction numerator k J over denominator m o l cross times K end fraction close parentheses and a normal boiling point of 80.1oC. At what temerature, in oC, does benzene boil when the external pressure is 895-torr?
Explanation / Answer
1) No of mol of ethylene glycol = 40.89/62.07 = 0.66 mol
No of mol of water = 677/18 = 37.61 mol
mole percent (%) of ethylene glycol = 0.66/(37.61+0.66)*100 = 1.72%
mole percent (%) of water = 37.61/(37.61+0.66)*100 = 98.27%
2) no clear information
3)
heat of vaporisation DHrxn = 30.72 kj/mol
normal boilingpoint T1 = 80.1 c = 353.25 k
atmospheric pressure P1 = 760 torr
external pressure P2 = 895-torr
boiling point at P2 = T2 = ?
from clasius-cleperon equation
ln(p2/p1) = DHrxn/R[1/T1 - 1/T2]
R = gas constant = 8.314 j.k-1.mol-1
ln(895/760) = (30.72*10^3/8.314)((1/353.25)-(1/T2))
T2= 358.85 k = 85.7 C
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