Question

Which of these aqueous solutions would be expected to have the highest boiling point?

a. 0.100 m CaCl2

b. 0.200 m NaOH

c. 0.050 m K2SO4

d. 0.050 m Al2(SO4)3

e. 0.200 m CH3OH

Answer 1

The solution with the highest boiling point is one that dissociates into the greatest number of ions. Option (d) 0.050 m Al2(SO4)3 should have the highest boiling point because it produces the most particles when dissolved in water, leading to the greatest boiling point elevation.

Step-by-step explanation:

To determine which aqueous solution would have the highest boiling point, we need to consider the colligative properties of solutions, particularly the boiling point elevation. The boiling point elevation is directly proportional to the concentration of the solute particles in the solution. Since ionic compounds dissociate into multiple ions in solution, their boiling point elevation effect is greater compared to molecular compounds, which do not dissociate. The van 't Hoff factor (i) represents the number of particles into which a solute dissociates in solution.

Using this knowledge, let's analyze the boiling point elevation for each solution given:

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Considering that the more particles in a solution, the higher the boiling point elevation will be, and thus the higher the boiling point of the solution will be, option (d) 0.050 m Al₂(SO₄)₃ should have the highest boiling point as it produces the most particles when dissociated.

Answer 2

0.200 m NaOH has the highest boiling point (option B)

Step-by-step explanation:

Step 1: Data given

The elevation of boiling point is proportional to the number of ions in solution.

The van't Hoff factor, i, is a measure of how many ions are produced from one molecule of the solute. For nonelectrolytes, i = 1).

We can compute the product of the van't Hoff factor and the molality for each of the solutes in the list:

a. 0.100 m CaCl2

⇒ CaCl ⇒ Ca^2+ + 2Cl- : van't Hoff factor = 3

0.100 * 3 = 0.300

b. 0.200 m NaOH

⇒ NaOH ⇒ Na+ + OH- : Van't Hoff factor = 2

0.200 * 2 = 0.400

c. 0.050 m K2SO4

⇒ K2SO4 = 2K+ + SO4^2- : Van't Hoff factor = 3

0.050 * 3 = 0.150

d. 0.050 m Al2(SO4)3

⇒ Al2(SO4)3 ⇒ 2Al^3+ + 3SO4^2- : Van't Hoff factor = 5

0.050 * 5 = 0.250

e. 0.200 m CH3OH

methanol is a non-emektrolyte Van't hoff factor = 1

0.200 * 1 = 0.200

Since the highest result is from 0.200 m NaOH, means this has the highes boiling point. ( Option B) is correct)