(a) A solution of 2.000 g of the compound Y(CH6N2O5) dissolved in 40.00 g water has a freezing point of -2.38 degree C. What is the molar mass of the compound Y(CH6N2O5) ? what is the implication of this molar mass? (Given that the freezing point depression constant Kf of water is 1.86 degree C/m.)

Answer 1

And
(b) Predict which would e cheaper for preparing an antifreeze solution, after adding the following solutes (given that they sold at the same price per pound): 0.15 m Pb(ClO3)2 or unknown compound Y(CH6N2O5) stated above? Answer:_________
Why?

Answer 2

dT = Kf*m
2,38 = 1.86*m
m = ?
Then m = #mols/kg solvent
m = #mols/0.04 = ?m
#mols = grams/molar mass
# mols = 2.00/molar mass
Solve for molar mass. I have approximately 39
With a formula of CH6N2O5 we're looking at about 126. The implications is that compound Y must not be ionized to some degree. For example:
If it were ionized into two particles then
2.38 = i*Kf*m. Go through the calculations and
molar mass = 78 etc

Answer 3

ok, I also counted 39 in (a)
actually I am so sorry that I tape a wrong compound, compound Y should be CH6N2O2 , will it have same explain in (b)?

Answer 4

Yes and no and I wondered about that when I responded to the initial question. In this case note that the CH6N2O2 adds up to 78 which is exactly twice the 39 we obtained for the molar mass. That means that i in
dT = i*Kf*m is 2 so the compound has ionized into two particles. This makes a lot more sense.