# Question 8933a

Apr 3, 2017

No, you are not correct.

#### Explanation:

Freezing-point depression is a colligative property, which means that it depends on the number of particles of solute present in solution, not on the nature of said particles.

In other words, the bigger the number of particles of solute present in solution, the greater the freezing-point depression and the lower the freezing point of the resulting solution.

In your case, the solutes are

• potassium sulfate, ${\text{K"_2"SO}}_{4}$
• copper(II) oxide, $\text{CuO}$
• dinitrogen trioxide, ${\text{N"_2"O}}_{3}$
• calcium oxide, $\text{CaO}$

The greatest decrease in the freezing point of the solution will be caused by the solute that produces the biggest number of particles when dissolved in water.

Dinitrogen trioxide is a covalent compound, so it will dissolve without dissociation. This implies that for every $1$ mole of dinitrogen trioxide you dissolve in water, you get $1$ mole fo particles of solute.

Copper(II) oxide is insoluble in aqueous solution, which means that it will hardly affect the freezing point of the solution.

Calcium oxide will react with water to form calcium hydroxide

"CaO"_ ((s)) + "H"_ 2"O"_ ((l)) -> "Ca"("OH")_ (2(s))#

Calcium hydroxide is a strong base, which implies that it dissociates completely to form calcium cations and hydroxide anions, but it is not very soluble. So once again, the freezing point of the solution will not decrease significantly.

Finally, potassium sulfate is a soluble ionic compound, so it will dissociate completely in aqueous solution to form potassium cations and sulfate anions

${\text{K"_ 2"SO"_ (4(aq)) -> 2"K"_ ((aq))^(+) + "SO}}_{4 \left(a q\right)}^{2 -}$

Notice that for every $1$ mole of potassium sulfate that dissolves in water, you get $3$ moles of particles of solute

$\text{2 moles K"^(+) + "1 moles SO"_4^(2-) = "3 moles ions}$

This lets you know that out of all the solutes listed, potassium sulfate will cause the greatest decrease in the freezing point of the solution.