Archive for October 14th, 2008
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A 650 ml aqueous solution contains 70 g of urea CO(NH2)2. The vapor pressure of pure water at 20.0 degrees C is 17.5 mm Hg. What is the vapor pressure of the water in the urea solution? (assume that the molar concentration of water in this solution is 55.5 M)
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Moles urea = 70 g /60.07 g/mol =1.17
Moles water = 55.5 x 0.650 L =36.1
Mole fraction water = 36.1 / 36.1 + 1.17 =0.969
p = p°X = 17.5 x 0.969 = 17.0 mm Hg
October 14th, 2008
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At 20˚C the vapor pressure of benzene is 75 torr, and the vapor pressure of toluene is 22 torr.
(a) A solution is prepared from 1.0 mole of biphenyl, a nonvolatile solute, and 49.0 moles of benzene. How do I calculate the vapor pressure of the solution at if the temp. is at 20˚C.
(b)A second solution is prepared from 3.0 moles of toluene and 1.0 mole of benzene. How do I determine the vapor pressure of this solution and the mole fraction of benzene in the vapor?
Chemistry Tutor Site - Chemistry Homework Answer
(a) From Raoult’s law: Vapour pressure = mole fraction in liquid * SVP
For benzene:
Vapour pressure = 49/(1+49) * 75 = 73.5 torr
The biphenyl is involatile so its vapour pressure is 0 torr.
Vapour pressure of solution = 0 + 73.5 = 73.5 torr
b) Using the same equation:
For toluene:
Vapour pressure = 3/(3+1) * 22 = 16.5 torr
For benzene:
Vapour pressure = 1/(3+1) * 75 = 18.75 torr
Vapour pressure of solution = 16.5 + 18.75 = 35.25 torr
Now, using Dalton’s law:
Partial pressure = mole fraction in vapour phase * total pressure
Mole fraction = Partial pressure / total pressure = 18.75 / 35.25 = 0.532
October 14th, 2008
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What is the freezing point of radiator fluid that is 50% antifreeze by mass? Kf for water is 1.86 degree C/m.Ethylene glycol, the primary ingredient in antifreeze, has the chemical formula C2H6O2. The radiator fluid used in most cars is a half-and-half mixture of water and antifreeze.
Chemistry Tutor Site - Chemistry Homework Answer
n 100 g there are :
50 g of water = 0.050 Kg
and 50 g of C2H6O2 => 50 g / 62 g/mol = 0.806 moles
m = 0.806 / 0.050 Kg =16.1
delta T = 16.1 x 1.86 =30 °C
Thus the freezing point of this solution is - 30 °C
October 14th, 2008
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A student makes a solution that made from 135 g of Calcium Chloride in 155 mL of water? Can you calculate the freezing point temperature of that solution ?
Chemistry Tutor Site - Chemistry Homework Answer
Moles of caCl2
= mass/Mr
= 135/111
= 1.216 mol
Molality of CaCl2
= mol/Kg
= 1.216 mol/0.155
= 7.85 m
freezing point depression:
= 1.86 deg/m x 7.85 m . 3
= 43.8030C
the “3″ is the Van Hoff factor of CaCl2 thus the freezing point is - 43.8030C
October 14th, 2008
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