Two beakers each contain 200 mL of water. Into one beaker, a student adds 5.0 g NaI. Into the other, she adds 5.0 g of KI. What is the expected effect on the boiling points of the solutions?A) the NaI solution will have a higher boiling point than the KI solutionB) the KI solution will have a higher boiling point than the NaI solutionC) the boiling points of both solutions will be elevated to the same temperature, above 100 degrees CelsiusD) the boiling points of both solutions will remain at 100 degrees Celsius

A) the NaI solution will have a higher boiling point than the KI solution

Explanation:

We must have it behind our minds that the addition of a solute to water elevates the boiling point of water.

Boiling point elevation is a colligative property. It depends on the amount of solute present.

We know that;

ΔTb = Kb m i

Where;

ΔTb = boiling point elevation

Kb = boiling point constant

m = molality

i = Van't Hoff factor

Since KI and NaI has the same number of particles(two particles each), the boiling point elevation (ΔTb) depends on the molality of each solution. The molality also depends on the molar mass of each substance. The molality of NaI is greater than the molality of KI hence the boiling point NaI is greater than the boiling point of KI.

So;

200mL of water = 200g of water

Mass of solvent = 200g/1000 = 0.2 Kg

Molality of NaI = 5g/150 g/mol * 1/0.2 = 0.167 m

molality of KI = 5g/166 * 1/0.2 = 0.151 m

Kb for water = 0.512 oC m-1

Boiling point of water = 100 oC

Let the boiling point of NaI be A

A - 100 = 0.512 *  0.167  * 2

A = (0.512 *  0.167  * 2) + 100

A = 100.171 oC

Let the boiling point of KI be B

B - 100 = 0.512 * 0.151  *  2

B = (0.512 * 0.151  *  2) + 100

B = 100.154 oC

Hence;

Boiling point of NaI > Boiling point of KI

4d^55s^1 ( third choice)