Calculate the freezing point of a solution containing 25 grams of kcl and 2750.0 grams of water. the molal freezing point depression constant (kf) for water is 1.86 ∘c/m.1.23 oc-0.45 oc+0.45 oc-0.23 oc+0.23 oc

-0.45 °C

Explanation:

The freezing point depression can be calculated using the formula:

Where ΔT is the change of the freezing point temperature (from 0°C of pure water), Kf is 1.86 °C/m, b is the molality, and i is 2 (in this case with KCl).

All data required is given except for b, so we calculate the molality:

25 g KCl ÷ 74.55g/mol = 0.3353 mol KCl

2750 g water ⇒ 2750/1000 = 2.75 kg water

Now we calculate ΔT:

So the freezing point of the solution is [0°C - 0.45°C] = -0.45 °C

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.