Chemistry 122 [Tyvoll]
Fall 2008
Solutions for Practice Problems Using Molality and Colligative Properties
1. What is the molality of a sucrose solution that contains 582 g of sucrose (a nonelectrolyte)
dissolved in 687 g of water? Sucrose: 342.3 g/mol
m = (582 g of sucrose/687 g H2O)(1 mol sucrose/342.3 g sucrose)(103 g H2O/kg H2O)
m = 2.47 mol sucrose/kg H2O = 2.47 m
2. How would you prepare 800 g of a solution that is 6.25 m ethylene glycol (a nonelectrolyte)?
Ethylene glycol: 62.07 g/mol
6.25 m = (6.25 mol eg/103 g H2O) = (6.25 mol eg)(62.07 g eg/mol eg)/103 g H2O)
6.25 m = 387.9 g eg/103 g H2O)
g soln. = 1000 g + 387.9 g = 1387.9 g @ 387.9 g eg/1387.9 g soln. = x g eg/800 g soln.
x = 223.6 g eg
800 − x = 800 g – 223.6 g = 576 g H2O
Dissolve 223.6 g ethylene glycol in 576 g H2O.
3. What is the freezing point for a solution that is 0.25 m urea (a nonelectrolyte) if the normal
freezing point of water is 0.000 0C and the molal freezing point depression constant for water
is Kf = 1.86 0C/molal? @ ΔTf = Kfm
So, ΔTf = Kfm = (1.86 0C/molal)( 0.25 molal) = 0.465 0C
t(soln) = t(H2O) −ΔTf = 0.000 0C − 0.465 0C = − 0.465 0C
4. A 2.216 g sample of coniferin, a sugar derivative found in fir trees was dissolved in 48.68 g of
water and had a boiling point of 100.068 0C. What is the molar mass of coniferin if we assume
Kb = 0.512 0C/molal (given)
that this compound is a nonelectrolyte? @ ΔTb = Kbm
ΔTb = Kbm = 100.068 0C − 100.000 0C = 0.068 0C = (0.512 0C/molal)(m)
m = 0.068 0C/(0.512 0C/molal) = 0.1328 molal
molar mass = M = (2.216 g coniferin/48.68 g H2O)(1000 g H2O/0.1328 mol coniferin)
M = 3.42 x 103 g coniferin/mol coniferin