Can anyone help me with this? I don't understand it. It's Colligative properties.?

Question

Step by step instructions with answers please. I need to check if I'm following it correctly.

What is the freezing point depression of 17.1 g sucrose and .10 water?

If 0.50 mol of Nacl are dissolved in 500.0g of ether, what is the freezing point of the solution?

If 222 g CaCl2 are dissolved in 1000 g of water what is the freezing point of the solution?

What is the boiling point elevation of 74.5 g KCl and 400.0 g water?

What is the boiling point of a solution of 149.0 g KCl and 400.0 g water?

A solution contains 450.0 g sucrose and 125.0 g water. What is the boiling point of the solution?

Thanks for your help.

Answer 1

The change in the freezing point is

ΔTf = i*Kf*m

Kf is a constant which depends on the solvent
m is molality= moles of solute per kg of solvent. Based on the data you have m=1000*mass solute/(MW solute* mass solvent)
e.g. for the first sucrose has MW=342.3 so m=1000*17.1/(342.3*0.10)

i is the Van't Hoff coefficient. You can consider that its value represents the number of particles that form when one molecule of solute is dissolved in the solvent
For non-electrolytes i=1 since the molecule will remain as it is. Sucrose is not an electrolyte so it has i=1
For strong electrolytes i= the number of ions formed when 1 molecule dissolves. So for NaCl and KCl i=2 (1 ion Na+ or K+ and 1 Cl-), for CaCl2 i=3 (1 ion Ca+2 and 2 Cl-)

HOWEVER NaCl will NOT dissociate in ether, since ether is not a polar solvent. Actually it should barely dissolve, but anyway... Thus, while it has i=2 in water, it has i=1 in ether.(I think that's the purpose of this question; to help you understand how to use i)

For the elevation of the boiling point you have similarly
ΔTb = i*Kb*m

The freezing point of the solution will be Tfsolvent-ΔTf and the boiling point Tbsolvent+ΔTb.
You need to find the values of Kf,Kb for the solvents you have (they should be given in your textbook) and the freezing point of ether.

Answer 2

Colligative properties are those that depend on the molality of the solution: The number of moles of solute dissolved in 1 kg of solvent. This is different from molarity. Your first question I do not understand, because .10 water is meaningless. The second question is astounding, because one cannot have 0.50 mol of an ionic salt NaCl dissolved in ether. Even if that were possible, you do not give the molal freezing point depression constant of ether. You will do better to ask the rest of your questions fewer at a time. That way, you will get more people to work on them.

Answer 3

Noone is going to do all that work for you...
Here's the right formulas:

dT =freezing point depression/boiling point elevation

dT=(R*T[f/b]^2*M1*m) /(1000* dH[vap/fus])
or dT =K[b/f]*m
These are known as the molal boiling point elevation/freezing point depression formulas.

T[f/b] =boiling/freezing point of solvent
M1 = molecular wieght of solvent
m = 1000*(n/w) = 1000*(moles of solute/grams of solvent)
dH[vap/fusion] = heat of vaporization/fusion for solvent