Quote (saber_x3 @ Jul 1 2015 05:21pm)
This is the same as freezing point depression, just using the boiling constant instead of freezing
It doesn't give me grams, nor moles, nor anything of anything though.
If you dissolve of NaF in water and find that the new boiling point of the solution is 103.1OC. What is the mass percent of NaF?would this be 3.1 = (2) (m) (.5121)
and if I solve for m I get 3.026752587
This post was edited by chicano on Jul 1 2015 05:30pm