freezing point?

Question

61.)
what is the freezing point of a solution that contains 0.5 moles of Nal in 500g of water? (Kf = 1.86C/m; molar mass of water = 18g)

please help and explain this to me??
i have a test 2mrow

<3

Answer 1

mathematical expression for Freezing Point Depression:

Delta Tf = Tf0 - Tf= i Kf m

where:

Tf = Freezing Point of the solvent in the solution

Tf0 = Freezing Point of the pure solvent

i = i factor which will be 1 if the solute is a non-electrolyte

Kf = Freezing Point Depression Constant which should be negative

m = concentration in moles of solute per kilogram of solvent

Let's look at an application using the above equation.

Calculate the Freezing Point of a solution of 90 grams of glucose (MW = 180) a non-electrolyte, dissolved in 750 grams of water. The Kf for water = -1.86 C/m. The Freezing Point of pure water = 0 C

1. Convert 90 grams into moles of glucose by dividing the grams by the molecular weight of glucose (180).

90 grams glucose X 1 mole glucose / 180 grams glucose = 0.5 moles glucose

2. Convert grams of solvent water to kilograms of water by dividing the grams by 1000

750 grams H2O X 1Kg / 1000 grams = .750 Kg H2O

3. Calculate the molal concentration using the definition for molal.

m = moles of solute / kilograms solvent = 0.5 moles glucose / .75 Kg H2O = 0.667 molal

4. Calculate the Freezing Point of solution by using the Freezing Point Depression Equation

0 - Tf = (1.86) (1) (.667) = 1.24

Tf = -1.24 C

Answer 2

Freezing points are depressed by the addition of a solute. The amount of depression is based on the number of g-ions (for an ionic substance) or g-moles in solution. Kf is per amount of ions or moles per 1000 g of water.

So your 0.5 moles of NaI will create 1 g-ion in the 500 g solution. Scaling up to 1000 g, this would be 2 g-ions. So the depression is 1.86 x 2 = 3.74 degrees; the freezing point is -3.74 degrees C.