A 0.227M solution of X (M
m
= 195 g/mol) has a density of 0.959 g/ml. What is the
molality of X in the solution?
Given Concentration Unit (GCU):
0.227 mol X / 1 L of soln
Desired Concentration Unit (DCU):
mol X / kg of solvent
Note: we have mol X in both, so problem is effectively that of converting 1 L solution to
mass of solvent.
1.
Since GCU is expressed per L of solution, a convenient “size” sample is 1.0 L.
Thus, 0.227 mol X.
2.
The only information we have about the solution is the density, allowing us to
convert volume of soln to mass soln.
1L * (0.959g/ml)*(1000mL/1L) = 959 g or
0.959 kg of solution.
3.
mass of solution = 0.959 kg = mass of solvent + mass of solute .
Since we need
kg of solvent in the DCU, we must obtain mass of solute.
4.
The only information given about the solute is the molar mass.
As we have mol
X, we convert:
0.227 mol X * (195 g/mol) * (1kg/1000g) = 0.0443kg solute.
5.
Mass of solvent = 0.959 – 0.0443 = 0.915 kg.
6.
Molality = 0.227 mol X/ 0.915 kg solvent = 0.248 m.
Colligative Properties:
P. Exp 2.14, p. 53.
62.3 g CaCl2 in 100 mL of water at 20ºC.
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 Fall '07
 Warren
 Chemistry, Physical chemistry, Vapor pressure, Freezingpoint depression

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