Help please. How do i begin this? A solution is prepared by dissolving 12.6g of HCl in 225g of water. Calculate the freezing point of the solution assuming complete ionization of the HCl in water.

Water: mp = 0.0 degrees C
kf = 1.86 degrees C/m

Question

Help please. How do i begin this? A solution is prepared by dissolving 12.6g of HCl in 225g of water. Calculate the freezing point of the solution assuming complete ionization of the HCl in water.

Water: mp = 0.0 degrees C
          kf = 1.86 degrees C/m

matt101 · Answer

You will need the equation for freezing point depression:

$$\Delta T_{f} =imK_f$$
The CHANGE in freezing/melting point is equal to the van't Hoff factor times the molality times the freezing point depression constant (specific to the solvent). We already have Kf, but we need to find i and m.

The van't Hoff factor, i, is the number of particles each solute molecule produces in solution. HCl dissociates into H+ and Cl-, so i = 2. Molality is the moles of solute per kilogram of solvent. Our solute is HCl, so we have 12.6/36.5 = 0.345 mol of HCl, and we have 225 g or 0.225 kg of our solvent, water. That means molality is 0.345/0.225 = 1.53 m.

Now we can plug these numbers into the first equation above to find the change:

$$\Delta T_{f} = 2 \times 1.53 \times1.86=5.69$$
This means the freezing point dropped by 5.69 C, so the new freezing point is 0 C - 5.69 C = -5.69 C.

If you have any questions please ask!

anonymous · Answer

Awesome! @matt101 Explanation was on point. Thank you very much!

matt101 · Answer

No problem - glad I could help!