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 = |0.376-0.410|= 0.034
% diff = (0.034/0.41)*100 = ?
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
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