Posted by SuSu on .
If 1030 g of a potassium citrate is dissolved in 2320 g of water solvent, calculate the freezing temperature (if below 0 oC, include the sign) of the solution. Consider if the solute is an electrolyte.
Kf (oC/m) 1.858

chemistry 
DrBob222,
This is done the same way but delta T = i*Kf*m
i will be 2 since potassium citrate dissociates into two ions. .
moles K citrate = grams/molar mass.
molality = moles K citrate/2.32 kg solvent.
Then plug into the first equation and calculate delta T. Use that to obtain the final T. 
chemistry 
SuSu,
is the answer 5.38???
this assignement is due in less than 20 min and i need an answer 
chemistry 
DrBob222,
Sorry, I went to bed last night. It was WAY past my bed time. I worked the problem and obtained delta T = 5.49 which is close to you 5.48. Probably just a rounding error BUT the answer is 5.49. Since delta T is 5.49, and the regular freezing point is 0, then delta T = 0T
5.49 = 0T
T = 5.49.