Question

. (1.5 pts) You dissolve HNO2 into water. HNO2 is a weak electrolyte. The molality of the HNO2 before any of it dissociates in the water is 0.200 m. The HNO2 is observed to dissociate 15.0% in the water. (a) What is the experimental molality of the HNO2 after it dissociates? (b) What is Tf? (c) At what temperature will the solution freeze?

Answer 1

a)

HNO2 dissociates as follows

HNO2 -----> H+ + NO2-

0.2-X.............X.......X

X = 0.2 x 0.15 = 0.03 (15 % dissociation)

Total Moles after dissociation = 0.2 - X + X + X = 0.2 + X = 0.2 + 0.03 = 0.23 moles

=> Experimental molality of HNO2 after it dissociates is 0.23 m

b)

We know that,

Kf = 1.86

Experimental molality = 0.23 m

delta Tf = Kf x m = 1.86 x 0.23 = 0.4278 degree C

c)

Temperature at which the solution will freeze = 0 - 0.4278 = -0.43 degree C