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Problem Set II
1. A. What is the vapor pressure of a solution prepared by dissolving 25.00 grams of ethyl
alcohol (molar mass = 46.0 grams/mole) in 100.00 grams (molar mass = 18.0 grams)of
water at 25 oC? The vapor pressure of pure water is 23.8 torr, and the vapor pressure
of ethyl alcohol is 61.2 torr at 25 oC? (Ans. = 27.2 torr)
b.
c.
d.
e.
f.
What is the mole fraction of water in the solution? (Ans. = 0.911)
What is the mole fraction of ethyl alcohol in the solution? (Ans. = 0.0890)
What is the vapor pressure of water above the solution? (Ans. = 21.7 torr)
What is the vapor pressure of ethyl alcohol above the solution? (Ans. = 5.45 torr)
What is the mole fraction of ethyl alcohol above the solution? (Ans. = 0.200)
2. What is the normal boiling point of a solution prepared by dissolving 1.50 grams of
aspirin (C9H8O4) in 75.00 grams of chloroform (CHCl3)? The normal boiling point of
chloroform is 61.7 oC, and Kb for chloroform is 3.63 oC/molal. (Ans. = 62.1 oC)
3. What is the freezing point of a solution prepared by dissolving 7.40 grams of K2SO4 in
110.0 grams of water? Assume 100% dissociation for K2SO4. The value of Kf for water
is 1.86 oC/molal. (Ans. = –2.15 oC)
4. What is the molality of an aqueous solution of KBr whose freezing point is –2.95 oC?
Assume complete dissociation for KBr. The Kf for water is 1.86 oC/molal. (Ans. = 0.793
molal)
5. A solution prepared by dissolving 20.0 grams of insulin in water and diluting to a
volume of 5.00 mL gives an osmotic pressure of 12.5 torr at 27 oC. What is the molar
mass of insulin? (Ans. = 5,987,098 grams/mole)
6. A solution is prepared by dissolving 10.085 grams of a non-ionizing compound in 50.0
grams of acetic acid. The boiling point of the solution is elevated by 1.76 oC.
a. What is the molality of the solution? (Ans. = 0.573 molal)
b. What is the molar mass of the compound? (Ans. = 352 grams/mole)
7. What is the vapor pressure of a solution that contains 10.0 grams of urea, CH4N2O, in
150.0 grams of water at 45 oC? The vapor pressure of pure water at 45 oC is 71.93 torr.
Ans. = 70.5 torr)
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