ΔTF = KF × b × i freezing point saltwater

[Gliese123]

Homework Statement

I think this is the right formula: ΔT_F = K_F × b × i
Freezing-point depression.
I have the amount of salt added to the water but how am I going to do to calculate the difference in temperature? Or with other words the freezing point of water saturated with salt (NaCl)? Or am I totally gone here? >_<
And K_F is 1.853 K·kg/mol.

But b and i, what should they include? Be nice, I'm an amateur :p

 

[Borek]

I guess you copied it from wikipedia (http://en.wikipedia.org/wiki/Freezing-point_depression#Calculation) - all symbols are explained there. Which one you have problems with?

Note that this equation works nicely for diluted solutions, but fails for concentrated ones. Saturated NaCl IS concentrated.

 

[Gliese123]

I guess you copied it from wikipedia (http://en.wikipedia.org/wiki/Freezing-point_depression#Calculation) - all symbols are explained there. Which one you have problems with?

Note that this equation works nicely for diluted solutions, but fails for concentrated ones. Saturated NaCl IS concentrated.

I don't really know the values to use. What should I use instead if concentrated ones doesn't work with that formula?

 

[Borek]

I don't really know the values to use.

They should be calculated. Although in this paritcular case...

What should I use instead if concentrated ones doesn't work with that formula?

...freezing point tables (determined experimentally) sound like a better solution.

 

[Gliese123]

Okay. Now let's say I want to know when salt water freeze when I've 1.5 g NaCl in 0.25 L water.
ΔT_F=1.853 * 0.025667 (mol NaCl) * 2 (ions Na⁺ & Cl^-)
= gives me 0.0951219. Which is totally wrong I think. It's not that little in temperature difference if I had salt into water.. Please explain...
-___- ' I would be very glad :)

 

[Borek]

Number of moles of NaCl is calculated OK, but moles are not the concentration.

http://www.chembuddy.com/?left=concentration&right=molality

 

[Gliese123]

Number of moles of NaCl is calculated OK, but moles are not the concentration.

http://www.chembuddy.com/?left=concentration&right=molality

Now I think I got it:
E.g. K_f=1.867 * 0.25/ 0.01388 * 2 * 0.1368= ~9.13 °C => Freezing point is reduced to -9.13 °C
I think it should be right? If b is (2g/58.44)/0.25= 0.1368
Which is 0.25 L water & 2 grams of salt. xD

 

[Borek]

Now I think I got it:
E.g. K_f=1.867 * 0.25/ 0.01388 * 2 * 0.1368= ~9.13 °C => Freezing point is reduced to -9.13 °C
I think it should be right? If b is (2g/58.44)/0.25= 0.1368
Which is 0.25 L water & 2 grams of salt. xD

I have no idea what is what of what in what you wrote - but it is wrong.

This is a simple plug and chug. The only thing you have to do is to correctly calculate concentration. You don't have to rearrange the equation to calculate K_f, as K_f is given, besides, K_f is not what you are looking for - ΔT_F is.