A solution is made by dissolving 10.20 grams of glucose (C6H12O6) in 355 grams of water. What is the freezing-point depression of the solvent if the freezing point constant is -1.86 °C/m? Show all of the work needed to solve this problem.

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anonymous · Answer

Well, freezing point of pure water is 0°C, with that in mind you can use $$DeltaTf = i(Kf*m)$$
To work out the change in freezing point, then all you have to do is subtract that change from the initial freezing point of zero. 
 "i" is the van Hoff't factor - and indicates the effective number of ions in the solution. ( In this case its taken as one, as glucose is not volatile).

You already have Kf (its the freezing point constant) and "m" is molality, which is mol solute (glucose) over kilograms of solvent. 

So all you need to do is work out the mol of solute, convert grams of water to kilograms. Plug that into the equation for change in temp and then subtract that from the initial freezing point.