Please help me with Freezing Point one last time? I am still having trouble and I only have one last try to get it correct? Where am I going wrong?

Please solve in steps and show all work.

A solution of sodium bromide (NaBr) is prepared by dissolving 6.24 g of the salt in 1.00 kg water (H2O). What is the freezing point of this solution? (The freezing point of pure water is 0°C (exactly) at 1 atm pressure; Kf= 1.86°C/m.

I have worked on this problem for 45 mins now. 0.2256 is incorrect ... Please Help!

1 Answer
Feb 4, 2018

Have you accounted for the speciation...?

Explanation:

By which I mean the reaction of sodium bromide with water....

#NaBr(s) stackrel(H_2O)rarrNa^+ +Br^-#

TWO MOLES of particles result from the one mole of salt.

#"Molality of NaBr"=((6.24*g)/(102.89*g*mol^-1))/(1*kg)=0.061*mol*kg^-1#...

And so WITH RESPECT to the particles in solution we have a molality of #0.121*mol*kg^-1# (NB #"molality"="moles of solute"/"kilograms of solvent"#.

And thus .................

#Delta_T=0.122*mol*kg^-1xx1.86*""^@C*mol^-1xx1*kg=0.227# #""^@C#...

And so the freezing point of the solution is approx. #-0.23# #""^@C#.

Good luck, and please give us some feedback....