What is the freezing point of an aqueous solution that boils at 105.9 ∘C? Express your answer using two significant figures.

THE FREEZING POINT OF THE AQUEOUS SOLUTION IS - 7.3 °C

Explanation:

To solve this problem, we must know the following variables:

Normal boiling point of water (solvent) = 100 °C

The molar boiling point elevation constant of water  = 1.51 °C /m

Normla freezing point of water ( solvent) = 0 °C

The molar freezing point depression constant = 1.86 °C /m

The boiling point of the aqueous solution = 105.9 °C

Molarity = xM

Change in boiling point = boiling point of solution - boiling point of water

Change in boiling point = 105.9 - 100 °C

= 5.9 °C

From the formula:

Change in boiling point = i * Kb * M

Re- arranging the formula by making M the subject of the equation, we have:

M = change in boiling point / Kb

i = 1

M = 5.9 °C / 1.51 °C/m

M = 3.907 M

Then, we calculate the freezing point:

Change in freezing point = i * Kb * M

= 1 * 1.86 °C/m * 3.907 M

= 7.267 °C

Hence, the freezing point = freezing point of water - change in freezing point

Freezing point = 0 °C - 7.267 °C

Freezing point = - 7.267 °C

Freezing point = -7.3 °C

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