Please help me with this problem

Question

          

2.​Calculate the freezing point depression of a solution prepared by​dissolving 10.0g of Na2SO4 in 500g of water. At what temperature will this

​solution actually freeze.  

 

​Kf(H2O) = 1.86 degree/molal

(a) 0.786 oC  (b) -0.786 o C (c) 0.262 oC  (d) 0.524 oC (e) -0.524 oC

Answer 1

The equation you'll want to use for this is ∆T = imK

∆T = change in freezing point = ?

i = van't Hoff factor = 3 for Na2SO4 (2 Na+ and 1 SO42-)

m = molality = moles Na2SO4 / kg water = 10.0 g Na2SO4 x 1 mol / 142 g = 0.0704 mol/0.5kg = 0.1408 m

K = freezing constant for water = 1.86º/m

Solve for ∆T:

∆T = (3)(0.1408)(1.86) = 0.79º

Final freezing point of solution = 0.00º - 0.79º = -0.79ªC

Answer 2

So calculate the molality of the solution, using the definition of molality! Then the freezing point depression is = molality * Kf . "Depression" means the temperature is below the normal freezing point of water. If you have to look that up ....

-- Cheers, --Mr. d.