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.

@jim_thompson5910 do you do chem?

Question

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.

@jim_thompson5910 do you do chem?

anonymous · Answer

please guys, I need your help

anonymous · Answer

I'll give best answer & fan/medal to anyone. I really just need help understanding this.

anonymous · Answer

@SolomonZelman do you help with chem?

owlcoffee · Answer

Okay, after a quick read about the matter, This is what we call a "Colligative property" of a solution, basically when we add sugars or salts on a solution and these new particles we added actualy get in the way of the molecular forces to align the molecules into ordered lines, wich we know is a solid.

Freezing depression is a consecuence of what I just said, because of the molecular forces that are being interfered  the freezing point actually lowers.
Remember, a solid is formed when the particles have very low kinetic energy, wich is observed in ice.

The equation that allows me to calculate the change in temperature is:
$$\Delta  t =(K_f )(m)(i)$$
We have to know some little things first find the molarity and find Van't hoffs factor of that.

anonymous · Answer

how do we do that?

anonymous · Answer

@Owlcoffee