Question

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

Answer 1

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.

Answer 2

is the answer 5.38???
this assignement is due in less than 20 min and i need an answer

Answer 3

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 = 0-T
5.49 = 0-T
T = -5.49.