Name: _________________________________________
Date:____________________________
Homework 3
INSERT SPACES!!!! INSERT SPACES!!!
1.) The vapor pressures of two pure solvents, benzene (C6H6) and toluene (C7H8) at 25oC are 95.1 and
28.4 mmHg respectively. A solution is prepared that has equal mole fractions of benzene and
toluene (meaning the SAME X value for both!!). Determine the vapor pressures of benzene and
toluene and the total pressure above this solution. Consider the solution to be ideal. (hint: you might
want to review your gas law chapter to remember how to determine the total pressure)
Since Xa + Xb + Xc + Xd . . . = 1.00 and there are only two substances present:
Xbenz + Xtol = 1.00
If Xbenz = Xtol
Then 2X = 1.00
So X = 0.500
Pbenz = Xbenz x Popure benz
Pbenz = 0.500 x 95.1 mm Hg
Ptol = Xtol x Popure tol
Ptol = 0.500 x 28.4 mm Hg
Pbenz = 47.6 mm Hg
Ptol = 14.2 mm Hg
The vapor pressures of each went down (vapor pressure lowering!!) when the two solvents were mixed
together!
The total pressure will be the sum of each of the pressures of gases present:
Ptotal = Pa + Pb + Pc . . .
Ptotal = 47.6 mm Hg + 14.2 mm Hg =
61.8 mm Hg
2.) The vapor pressure of pure water at 20.0oC is 17.5 mm Hg. What is the vapor pressure above a
solution that contains 1.00 m urea at 20.0oC? (HINT: molal tells you the amount of water!!)
m
1.0
XH2O =
1.00 mole urea
therefore we have 1.000 kg of water!
1.0000 kg water
kg H2O x
1000 grams water 18.02 grams water
x
= 55.5 moles
1 mole water
1 kg
55.5
moles water
(1.00 mole urea + 55.5 mole water)
PH2O = XH2O x PopureH2O
PH2O = 0.982 x (17.5 mm Hg)
PH2O = 17.2 mm Hg
= 0.982
3.) The boiling point of a solvent is 78.5oC. What is the boiling point of the solvent if 10.0 grams of
aspirin (mm = 180.2 grams/mole) is dissolved in 250.0 gram of the solvent? (Kb of the solvent =
1.22oC/m)
If the boiling point is going to go UP – then 78.5 + 0.271 = new boiling point = 78.8oC
4.) Which should lower the freezing point of water more? 50.0 grams of NaCl or 50.0 grams of
Na3PO4? (prove your answer mathematically). If you had 5.0 m NaCl and 5.0 m Na3PO4 which
would lower the freezing point of water more? If the answers are different – explain why. (Kf of
water = 1.86oC/m)
50.0 grams of NaCl will lower the freezing point of water more.
If the concentrations (molalities) are the same:
Then the Na3PO4 will lower the f.p. of water more as the differentiating factor is the i
value. When the grams are used, the depression will depend on the molar mass of the
substance AND i. Since NaCl has a much smaller mass than Na3PO4, the molal amount
x i > than that for Na3PO4. But when the molalities are the same – the only factor that
matters is the i value