Calculate the freezing point of a solution containing 110.7 g of glucose, which is C6 H 1206 dissolved in 302.6 g of water. All right, so we're gonna say here that are freezing point of our solution equals the freezing point of our solution equals the freezing point of our solvent minus delta T f. Our solvent here is water. Pure water freezes at 0°C. So we need to do here. Is figure out what delta T. F is We're gonna say Delta t f equals I times k f times m glucose. Here's our salute. It is not Ionic. It's co violent, so it doesn't break up into ions. So I is one KF We're dealing with water as our solvent. Its freezing point constant is 1.86°C over morality. Then all we have to do now is find the moles of glucose and convert grams of water into kilograms of water kg to just move the decimal over three. 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. Mhm. We're gonna sit here for every one mole of glucose. Its mass is 180 0.156 g. And that's the weight of the six carbons. 12 hydrogen is and six oxygen's together. Grams, cancel out. And now I'm gonna have moles, which comes out to . Moles of glucose. So we plugged that here. So here these counselor with this morality 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 addition of glucose is negative three 78°C, so that will be our final answer.