Because urea is a molecular compound and does not dissociate when it dissolves, the dissolving of 1 mole of urea gives us one mole of particles in the solution. This keeps the problem a little simpler than it would be for an ionic solid.
K_fKf and K_bKb are the amounts by which the freezing point is lowered and the boiling point elevated (in that order) for each one mole of particles dissolved in the solution. This enables us to write:
1.86 xx C+0.52 xxC=51.86×C+0.52×C=5
where CC is the concentration of urea solution. The sum comes to 5 °C because that is the amount by which we wish to expand the liquid range.
Now, to solve:
2.38 xx C = 52.38×C=5
so C = 2.38/5 = 0.476C=2.385=0.476 moles/L is the necessary concentration.
Because we will prepare 0.100 L of the solution, we will require
0.476 "mol"/L xx 0.100 L = 0.04760.476molL×0.100L=0.0476 mol of urea.
Since the formula for urea is CH_4N_2OCH4N2O, its molar mass is 60.1 g/mol.
Therefore, the mass needed is
0.0476 xx 60 = 2.86 g0.0476×60=2.86g
