Which aqueous solution below is expected to have the lowest freezing point? Assume 100% dissociation for ionic species.
1)0.25m MgBr2
2)0.30m Na2SO4
3) 0.50m KCl
4) 1.0m sucrose
5) 0.40m Cr(NO3)3
I think that the answer is 0.30NaSO4 but I am not positive. Is this correct?
delta T = i*Kf*m
You want the largest delta T (which will give the lowest freezing point).
Kf is constant so we can forget that. The two that matter are i and m
So multily i*m for the 5 and the largest number wins. Remember i is the van't Hoff factor which is the number of particles produced when the materials are placed in solution. sucrose is l. KCl is 2, etc.
I understand everything except the van't Hoff factor part. I do not understand how you figure that out.
That how the material ionizes.
KCl ==> K^+ + Cl^- and i = 2
MgCl2 ==> Mg&2+ + 2Cl^- and i = 3
Na2SO4 == 2Na^+ + SO4^2- and i = 3
Cr(SO4)3 ==> Cr^3+ + 3SO4^2- and i = 4
Sugar doesn't ionize so it is just 1 particle
Sugar ==> sugar(aq) and i = 1
Ok so then I just have to find the mass of each solution
so KCl = 74.55 X 2
MgBr2 = 184.11 X 3
Na2SO4 = 94.05 X 3
Cr(SO4)3 = 340.21 X 4
So Cr(SO4)3 would have the largest delta T, so the lowest freezing point.
Is this correct?
Why do you need the molar mass?
delta T = i*Kf*m
You know i from my previous response and you know m from the problem. i*m that gives the largest number will be the one that gives the largest delta T.
Ok so do you mean 0.25MgBr2 X 3 = 0.75 ?
i equals 3 for MgBr2 correct?
So I think that 0.40m Cr(NO3)3 has the largest delta T. Would you agree?
To determine which aqueous solution is expected to have the lowest freezing point, we need to consider the concept of freezing point depression. Freezing point depression occurs when a solute is added to a solvent, reducing the freezing point of the solvent.
The extent of freezing point depression depends on the concentration of the solute particles, rather than the identity of the solute itself. In this case, we assume that all the solutes are fully dissociated when dissolved in water.
Now, let's consider each of the given options:
1) 0.25m MgBr2: This solution consists of one magnesium ion (Mg2+) and two bromide ions (Br-) per formula unit of MgBr2, so the concentration of particles is 0.25 × 3 = 0.75 m.
2) 0.30m Na2SO4: This solution contains two sodium ions (Na+) and one sulfate ion (SO42-) per formula unit of Na2SO4, so the concentration of particles is 0.30 × 3 = 0.90 m.
3) 0.50m KCl: This solution consists of one potassium ion (K+) and one chloride ion (Cl-) per formula unit of KCl, so the concentration of particles is 0.50 × 2 = 1.00 m.
4) 1.0m sucrose: Sucrose is a non-ionic compound, meaning it does not dissociate into separate ions when dissolved. Therefore, its concentration is equal to its given value, 1.0 m.
5) 0.40m Cr(NO3)3: This solution contains one chromium ion (Cr3+) and three nitrate ions (NO3-) per formula unit of Cr(NO3)3, so the concentration of particles is 0.40 × 4 = 1.60 m.
Now, comparing the concentrations of solute particles for each option, we find that the solution with the lowest concentration of solute particles is 0.25m MgBr2 (0.75 m).
Hence, the correct answer is option 1) 0.25m MgBr2 is expected to have the lowest freezing point.