A solution of ethanol in water boils at 101.5 degrees C. what is the molarity?
delta T = i*Kb *molality.
Delta T = 101.5 - 100 = 1.5. i = 1. I should comment here that the freezing point constant and the boiling point contant works with a non-volatile solute in a solvent such as water. In your first problem you had glucose which is a non-volatile solute. Same for CaCl2; however, ethanol is a volatile solute and technically this solution doesn't follow the same rules. I think the spirit of the problem, though, is to work this problem using the same kind of information. The boiliing point constant for water is 0.51 degrees C/m.
okay this equation is for the C6H12O6 one....
2.50=1.86m you will divide and get
m=1.344 then you change that into grams and that is your answer??
I worked this in detail at the original post.
what???
Go to the original post to see the C6H12O6 problem worked in detail.
okay!!! i think i got it thanks!!