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.
Explanation:
It is given that volume of 
is 300 mL and molarity is 0.200 M.
Volume of NaCl is 200 mL and molarity is 0.050 M.
The chemical reaction will be as follows.
for 
is given as 
.
As, molarity is number of moles present in liter of solution.
Hence, moles of 
(aq) will be calculated as follows.
moles of (aq) = 
= 0.06 mol
![[Pb^{2+}]](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D)
= 
= 0.120 M
Mole of 
= 
= 0.010 M
Now, Q = ![[Pb^{2+}][Cl^{-}]^{2}](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D%5BCl%5E%7B-%7D%5D%5E%7B2%7D)
= 
= 
As, Q < hence, there will be no formation of precipitate.
What answers are there? you didn't add them.
physical change and chemical change.
physical change affects a substance's physical properties, and a chemical change affects its chemical properties.
I think mm is not a unit of pressure
