A solution is made by dissolving 15.5 grams of glucose (C6H12O6) in 245 grams of water. What is the freezing-point depression of the solvent if the freezing point constant is -1.86 °C/m?

Question

anonymous · Answer

Do you know the equation that can be used to solve this?

anonymous · Answer

No i do not

anonymous · Answer

If you give it to me I'm sure i can figure it out

anonymous · Answer

It is DeltaTf = Kfm; Delta Tf = change in freezing point, Kf = the freezing point constant, and m = moles solute/kg solvent (molarity).
So, you are given Kf, and you need to find m and DeltaTf.

anonymous · Answer

Remember to convert to the applicable units (i.e. kg instead of g.)

anonymous · Answer

Also, first identify which substance mentioned is the solvent and which is the solute.

anonymous · Answer

Does this make sense?

anonymous · Answer

Sorry, m = molality not molarity

anonymous · Answer

molality is moles of solute divided by kg of solvent

anonymous · Answer

ok i think im understanding

anonymous · Answer

So, convert H2O and glucose to kilograms.

anonymous · Answer

wait what is the -1.86 suppose to stand for

anonymous · Answer

That is Kf, the freezing point constant.

anonymous · Answer

ok then thank you