# A solution made by dissolving 10.20 g glucose C6H12O6 in 355 g of water.What is the freezing-point depression of the solvent if freezing point constant is -1.86 I don't understand this one /:

Remember, that was we add solute to water we lower the freezing point of the solution compared to the freezing point of pure water.  The equation we use to determine this is

deltaT = i Kf m

delta T is the change in the freezing point of the solvent minus the solution.  i is the van't Hoff factor that tells us how many particles we will get from each unit of solute. Kf is the freezing point depression constant which is characteristic of the solvent.  m is the molality of the solute in the solution (moles of solute/kg of solvent).

The first thing we need to do is find the moles of glucose in 10.20 g.   Since the molar mass is 180.2 g/mol, we can divide 10.20 into 180.2 and find 0.0566 mol of glucose.  Now that we've done that, we can start plugging in what we do know.  Since glucose does not break apart in water (as an ionic compound would), the i value for glucose is 1.

deltaT = 1 (-1.86)(0.0566 mol/0.355 kg)

deltaT = -0.297 degC

Freezing point is -0.297 degC which is lower than the freezing point of water (0 degC).

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