Question

what is the molality of a water solution that freezes at -2.25 degrees celcious? water freezing point= -1.86 degrees celcious/m or water boiling point=0.52 degrees celcious.

Answer 1

HELP

Answer 2

You would need to know the identity of the solute. molality is defined as moles of solute per kg of solvent. In this case the solvent is water, the solute ?
This is needed because from the information given this seems to be a freezing point depression / boiling point elevation problem. In that case the equation for freezing depression is: Freezing point depression = (van't hoff factor) x (molal freezing point depression constant, expressed as Kf) x molality.
The value of Kf for water is = 1.853 K*kg/mol
Van't hoff factor in a simplified form is the number of molecules that the solute dissolves. For example, for NaCl it is 2 because in solution it splits into Na+ and Cl- atoms. But for molecules that do not dissolve like sucrose, it is 1.

So as you can see you need to know the identity of the solute.

Answer 3

okay so how would i start off the problem?

Answer 4

Like I said since we do not know the identity of the solute we can't do this problem. But let's assume that the solute is sucrose. Since water normally freezes at 0 degrees celsius, but now it is freezing at -2.25 degrees celsius, the change (freezing point depression) is 2.25 degrees celsius. \[2.25 C = 1 \times 1.853 \frac{ C*Kg }{ mol } \times molality\]
And simply solve for molality

Answer 5

And this would give us a molality of 1.21 mol/kg

Answer 6

can you set up the problem ? like with the C and the Kg, mol, and molality

Answer 7

The C is degrees celsius. molality is just mol/kg. so as you can see the units cancel out on the right side, and you end up with Celsius on both sides.