Question #9cbbd

The thing to remember about the freezing point and the boiling point of a solution is that they depend on how many particles of solute are dissolved.

Dissolving a solute in pure water will produce a solution that has

• a boiling point that is higher than the boiling point of pure water
• a freezing point that is lower the freezing point of pure water

Now, the change in boiling and freezing point, i.e. the boiling-point elevation and the freezing-point depression, respectively, are proportional to something called the van't Hoff factor, #i#.

The van't Hoff factor tells you the ratio that exists between the number of moles of solute that are dissolved to make the solution and the number of moles of particles of solute that are produced in solution.

Simply put, the higher the value of the van't Hoff factor

• the higher the boiling-point elevation, i.e. the higher the boiling point of the solution
• the lower the freezing-point depression, i.e. the lower the freezing point of the solution

Now look at the three solutes given to you. You have

#"NaCl"_ ((aq)) -> "Na"_ ((aq))^(+) + "Cl"_ ((aq))^(-)#

Here #1# mole of sodium chloride dissolves to produce #1# mole of sodium cations and #1# mole of chloride anions, so

# color(blue)(ul(color(black)(i = 2))) -># one mole dissolved, two moles produced

#"CaCl"_ (2(aq)) -> "Ca"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-)#

Here #1# mole of calcium chloride dissolves to produce #1# mole of calcium cations and #2# moles of chloride anions, so

# color(blue)(ul(color(black)(i = 3))) -># one mole dissolved, three moles produced

#"Al"_ 2("SO"_ 4)_ (3(aq)) -> 2"Al" _ ((aq))^(3+) + 3"SO"_ (4(aq))^(2-)#

Here #1# mole of aluminium sulfate dissolves to produce #2# moles of aluminium cations and #3# moles of sulfate anions, so

# color(blue)(ul(color(black)(i = 5))) -># one mole dissolved, five moles produced

Therefore, you can say that dissolving aluminium sulfate in water will have the greatest effect on the boiling and freezing point of the solution.