Calculate the freezing temperature of a solution made by mixing 55.267 g of cacl2 (mw =110.98) in 250.00g of water
kf =1.86 Cm^-1 ( this is not centimetes)
So what I did was I looked at the problem in a plug and chug mindset, and the formula for freezing point is Tf=kf *m
BUT! CaCl2 is ionic which means we have to get the particals?
the equation will look like this 55.267 * 1mol/110.98 * 3/1
(1.86 C)( 1.49)/.350=7.94
0.00-7.94 = -7.94
My question is do you only get the particals for ionic?