Can someone explain how to do this?
a solution is made by dissolving 10.20 grams of glucose (c6h12o6) in 355 grams of water. what is the freezing-point depression of the solvent if the freezing point constant is -1.86 °c/m? show all of the work needed to solve this problem.

We use one concept of colligative properties which is freezing point depression. We solve it as follows:

ΔTf = kf (m)      where kf is the freezing point depression constant and m is the molality of the solution

m = n solute / m solvent = 10.20 g (1 mol / 180.18 g ) / .355 kg = 0.16 mol/kg

ΔTf = kf (m) 
ΔTf = -1.86 (0.16) = - 0.2976 °C

Hope this answers the question.

You could add six hundred seventy eight and three hundred three together by going right to left. three plus eight is eleven and so you carry the one and add the one with the seven and zero which is eight. finally, you add the six and three together which gives you nine. and you final answer is nine hundred eighty one
981