posted by Cody .
PLEASE SHOW PROCEDURE!
You wish to prepare an aqueous solution that has a freezing point of -0.100 degrees Celsius. How many milliliters of 12.0 M HCl would you use to prepare 250.0 mL of such a solution?
[Hint: Note that in a dilute aqueous solution, molality and molarity are essentially numerically equal.]
Your note gives away the solution:Shame on your teacher.
but m=M in dilute..
and for HCl you get two ions per molecule.
solve for M ( I get about .096)
so you want to dilute the HCl from 12 to .096, or dilute it 12/.096=125 times, which means you need one part HCl, 124 parts water
what is one part? 250ml/125parts=2ml
so measure out 248ml water, add 2 ml of the 12M HCl slowly (Always add acid to water).
check my reasoning.
This sounds like solid reasoning, but I apologize, because the answer in the back of my textbook says 0.56 mL of 12.0 M HCl.
I see your error. You used the incorrect freezing point depression constant: .52 instead of 1.86.