Check table T and use the concentration equation. Molarity= moles of solute/ liters of solution.
So 0.240 = x/ 1.65 once u find the # of miles of solute (x=.396) and substitute that
Wait I'm not sure if it's correct
I will have to go with carbon monoxide
I think it would be solubility but I’m not sure
This uses the concept of freezing point depression. When faced with this issue, we use the following equation:
ΔT = i·Kf·m
which translates in english to:
Change in freezing point = vant hoff factor * molal freezing point depression constant * molality of solution
Because the freezing point depression is a colligative property, it does not depend on the identity of the molecules, just the number of them.
Now, we know that molality will be constant, and Kf will be constant, so our only unknown is "i", or the van't hoff factor.
The van't hoff factor is the number of atoms that dissociate from each individual molecule. The higher the van't hoff factor, the more depressed the freezing point will be.
NaCl will dissociate into Na+ and Cl-, so it has i = 2
CaCl2 will dissociate into Ca2+ and 2 Cl-, so it has i = 3
AlBr3 will dissociate into Al3+ and 3 Br-, so it has i = 4
Therefore, AlBr3 will lower the freezing point of water the most.
32 gs, cuz 28 is an ounce and 3.5 is an 8th but if ur plugg blesses u he’s gon give u 32
