Okay, I have this problem and I can't figure it out!
What is the freezing point of a solution of 84.1 g of sodium chloride dissolved in 1000 g of water? The answer is -5.35 deg. I just dont know how to get the answer...thanks!
2007-07-27 08:49:10 · 3 answers · asked by Jay Jay 1 in Science & Mathematics ➔ Chemistry
A collogative property is one that depends on the total number of particles in a solution. NaCl will yield two particles (Na+ and Cl-) per molecule.
The freezing point depression is equal to the the moles of particles dissolved in 1000 gm of solvent (this is the molality) times a constant. The freezing point depression constant for water is 1.86 deg C per molatity.
84.1 g of NaCl is 1.439 moles of NaCl and 2.878 moles of particles.
2.878 x 1.86 = 5.35 deg C freezing point depression
2007-07-27 09:07:04 · answer #1 · answered by skipper 7 · 0⤊ 0⤋
The answer is given by the freezing point depression calculations that you will find in your chapter.
Calculate the molarity of the NaCl solution which you should be able to do easily. From there it falls in place
You need to master the material
2007-07-27 08:53:10 · answer #2 · answered by GTB 7 · 1⤊ 1⤋
Given ;
wt of solute = 84.1 g
wt of solvent = 1000 g
T1 = 0 ºC
Find ;
T2 =?
T2-T1 = kf(m )
T2 = kf ( m ) + T1
I can't solve this problem since kf is not given, pls refer to your chem books for kf values ( NaCl in water ) then you substitute the values on the above formula .
where m = molality
2007-07-27 10:43:21 · answer #3 · answered by Yheng Natividad 3 · 0⤊ 0⤋