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? Show all of the work needed to solve this problem

Question

matt101 · Answer

The equation for freezing point depression is:

$$\Delta T_f=iK_fm$$
Where ΔT(f) is the change in the freezing point, i is the van't Hoff factor, K(f) is the freezing point constant, and is the molality of the solution.

Do you think you can solve this question now?

anonymous · Answer

I think so... 
Um, if you're finding the molality, you do m of solute/kg of solvent. .17/.245=.69
Then you just plug everything in and multiply it. 
$$-1.86\times.69=-1.28$$ ???

anonymous · Answer

Don't you have to add the two now? Because when googled it, someone said to add them at the end.

anonymous · Answer

@matt101

aaronq · Answer

the formula is $\sf molality =\dfrac{moles~of~solute}{kg~of ~solution}$

not mass. you need to convert the mass of solute to moles first.

anonymous · Answer

okay, my bad, so the amount of moles is .09 i'm pretty sure. then when you divide it by .245 it's equal to .37. And then you would plug in a multiply all of them. So $$Delta T =-1.86\times.37$$ $$\Delta T =-.69$$

anonymous · Answer

@aaronq

matt101 · Answer

That's correct!