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