Question

A solution is made by dissolving 21.5 grams of glucose (C6H12O6) in 255 grams of water. What is the freezing-point depression of the solvent if the freezing point constant is -1.86 °C/m? Show all of the work needed to solve this problem.

Answer 1

I would like to understand how to do this fully please...

Answer 2

I'm guessing the freezing constant with unit C/m means celsius per molal?

Answer 3

I'm not sure but if so, would you just multiply the molality with that constant? Might wanna wait for someone else. But welcome to OpenStudy

Answer 4

I would divide the grams of glucose by the grams of water then multiply it by the constant. Im not too sure, correct me if im wrong.

Answer 5

Im not sure either..... would it look like this 21.5/255 x -1.8

Answer 6

yes, i would think so, but dont take my word though, im not sure....

Answer 7

okaaay. well thank u anyway :)

Answer 8

Well if I'm right, you find the volume of water using its density. Then divide the number of moles of glucose by the volume of water. Then multiply by the constant. That is, if I understand the constant correctly.

Answer 9

http://www.neurolab.ufsc.br/ensino/enq3248/HAYNIE2008_Biological_Thermodynamics.pdf

Answer 10

Try this @dwright25 I thought it is a good source for help

Answer 11

Oh you need more with that equation. Refer to the Calculation section http://en.wikipedia.org/wiki/Freezing-point_depression

Answer 12

thanks u guys

Answer 13

No problem, always available for help

Answer 14

use the formula: \(\Delta T=i*m*K_{\theta}\)

where, \(\theta=b\) when finding BP, and \(\theta=f\) when finding freezing point
(By that i mean use the appropriate constant.)

More on this: http://openstudy.com/study#/updates/521ecb3ae4b0750826e0c362