CHEMISTRY
posted by MAD on .
A) We have an experiment , using the solution of KNO3 , measured freezing point of solution is 1.15 degree , and using a sample of pure water the thermometer read o.25 degree as freezing point of the sample , calculate the molal concentration of KNO3 assume that the attraction force between the ions is ignored
B) If we do not ignore the attraction force , and decided to take a sample of that solution 10 mL ,, after boiling and vaporizing the water from the solution , got o.415 g of KNO3 , then find out the real molality of KNO3 , and what is percentage difference between measured concentration and real concentration of KNO3 , if the solution density is 1 g/mL
Thank you a lot

I think you meant to say that the freezing point of the pure water as read on your thermometer was 0.25. Therefore, the freezing point of the original solution KNO2 was 1.40.
Then 1.40 = i*Kf*m
1.40 = 2*1.86*m
m = about 0.376
0.415g KNO3 = 0.415/101.1 = about 0.0041 mols.
m = 0.0041mols/0.010kg solvent = about 0.410m
absolute difference = 0.3760.410= 0.034
% diff = (0.034/0.41)*100 = ?