Calculate the freezing point of a solution containing 110.7g of glucose, which is C₆H₁₂O₆ dissolved in 302.6g of water. All right, so we're going to say here that our freezing point of our solution equals the freezing point of our solution equals the freezing point of our solvent minus ΔT_F. Our solvent here is water. Pure water freezes at 0°C. So what we need to do here is figure out what ΔT_F is.

We're going to say ΔT_F = i × K_F × m. Glucose is our solute. It is not ionic, it's covalent, so it doesn't break up into ions. So i is 1. K_F, we're dealing with water as a solvent. Its freezing point constant is 1.86°C/m. Then all we have to do now is find the moles of glucose and convert grams of water into kilograms of water. Kilograms just move the decimal over 3.

So there goes our kilograms of our solvent water, and here we're going to find out the moles of our glucose. So we have this many grams of glucose. We're going to sit here for everyone. Mole of glucose. Its mass is 180.156g, and that's the weight of the six carbons, 12 hydrogens and six oxygens. Together grams cancel out and now I'm going to have moles, which comes up to 0.614467 moles of glucose.

So we plug that here. So here these cancel out. With this molality here, what I get at the end is 3.78°C, which is what I plug over here. So the new freezing point after the additional glucose is -3.78 degrees Celsius. So that would be our final answer.