A solution is prepared by dissolving 104.1 grams of barium chloride, BaCl₂, in 1000.0 grams of water. What is the boiling point of the solution? (The boiling point elevation constant for water is 0.512°C/mole solute in 1000 g of water.) A) 100.26°C B) 100.52°C C) 100.78°C D) 101.04°C

Question

Quizplus · Accepted Answer

First, we need to calculate the moles of barium chloride in the solution:

moles of BaCl2 = mass / molar mass = 104.1 g / (137.33 g/mol + 2*35.45 g/mol) = 0.5 moles

Next, we need to calculate the boiling point elevation:

ΔTb = Kb * molality

where molality = moles of solute / mass of solvent in kg

In this case, the mass of water is 1000.0 g, or 1.0 kg. The molality is:

molality = 0.5 moles / 1.0 kg = 0.5 m

The boiling point elevation constant for water is given as 0.512°C/mole. Therefore:

ΔTb = 0.512°C/mole * 0.5 m = 0.256°C

Finally, we add this to the normal boiling point of water (100°C) to get the boiling point of the solution:

boiling point = 100°C + 0.256°C = 100.78°C

Therefore, the best choice is C.

A Solution Is Prepared by Dissolving 104