how to calculate the molality?

Question

0.60 g of CCL4 in 430.0 g of benzene.

given information: BP of benzene = 80.1 C, Kb=2.53, Freezing pt=5.5 C, Kf = 5.12

i don't know if the information is necessary, but my book only shows how to calculate the molality of solute dissolved in water...

thanks

Answer 1

Molality is defined as the moles of solute divided by the kilograms of solvent.

So you are quite right; you don't really need to know the boiling point or freezing point to determine the molality.

First, you must convert your solute (.6 g of CCl4) into moles. You need to know the molar mass of CCl4 in order to do this. To determine the molar mass of a compound, simply add up the masses of the atoms that make it up: 12 + (35.5 * 4) = 154 g/mole.

To convert, use basic stoichiometry:
.6 g * (1 mol/ 154 g) = .004 moles

Then divide this by the kilograms of solvent ( .43 kg):
.004/ .43 = .009 = molality

Answer 2

Molality is the same without water.

The solute is the smaller, the solvent is the larger.

molality = moles of solute / kg. of solvent

convert the grams of CCl4 to moles. The benzene is 0.430 kg. of solvent

Answer 3

the correct definition of molality is moles of solute [CCl4] per kilograms of solvent [benzene].

moles of solute: CCl4 molecular weight is 153.82g; so 0.60 g of CCl4 is 0.60g/153.82 or 3.9e-3 moles.

kilograms of solvent: 430.0g of benzene is 4.3e-1 kg of benzene.

the molality is the ratio 3.9e-3mol/4.3e-1kg benzene

Answer 4

I have always defined Molality as moles/liter so.. you need the specific gravity of benzene..not the other stuff..

carbon tetra chloride has how many moles per gram? (12 plus 4 x 35)

0.6 grams / (12+4x35) = MOLES OF CARBON TET

430 grams of benzene x specific gravity = ml of benzene

(moles of carbon tet/ml)*(1000 ml/liter) = molality