Posts filed under 'Colligative Properties of Solution'
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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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A student in the laboratory weight a 2.11 gram sample of naphthalene (C10H8) dissolved in 35.00 grams of paraxylene that has a freezing point of 11.25 degrees celcius. The pure solvent has a freezing point of 13.26 degrees celcius. What is the freezing point depression constant of paraxylene?
Chemistry Tutor Site - Chemistry Homework Answer
Here are the formula
dT = kf x m x i
dT is change in temperature = 13.26C - 11.25C = 2.01 C
kf = freezing point depression constant for the solvent
m = molality
= moles solute / kg solvent
= (2.11 g / 128.2 g/mole) / 0.035 kg
= 0.0165 / 0.035
= 0.470 molal
i = van’t hoff factor = number of ions the solute dissolves into = 1 for a compound that doesn’t ionize (NaCl, for example has i = 2, Na2CO3 has i = 3)
rearranging….
kf = dT / (m x i) = dT / m = 2.01 C / 0.470 m = 4.27 C / m
October 14th, 2008
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I’m trying to find out the boiling point of a liquid using the Clausius-Clapeyron equation. I have no problem determining pressure or Hvap, but I think I’m rearranging it wrong when it comes to the temperature. Thanks for your help.
Chemistry Tutor Site - Chemistry Homework Solution
If you have the Hvap then you have your answer (or as close as you may get). The Clausius-Clapeyron equation explains the boiling point (vaporization) as a difference between 2 points.
To determine the boiling point of an unknown use a different point of the equation:
ln[(P2)/(P1)] = [(∆Hvap)/R]*[(1/T1)-(1/T2)]
Use the ln p vs 1/T graph to find the ln p and 1/T values of a point on the straight line. Let these values be equal to ln p and 1/T . You have already determined ∆ Hvap in this experiment. T2 will then be the normal boiling point occuring at p2 = 1 atm..
October 14th, 2008
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