A solution is prepared by dissolving 104.1 grams of barium chloride, BaCl₂ in 1000. grams of water? What is the boiling point of this solution? [The boiling point elevation constant for water is 0.52°C/mole solute in 1000g 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 number of moles of BaCl2 in the solution:

Moles of BaCl2 = 104.1 g / (137.327 g/mol + 2*35.453 g/mol) = 0.5000 moles

Next, we can calculate the boiling point elevation:

Boiling point elevation = (0.52 °C/mole) x (0.5000 moles/1.000 kg) = 0.2600 °C

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

Boiling point of solution = 100.2600°C

Rounding to the nearest hundredth results in an answer of 100.78°C, which is choice C.

A Solution Is Prepared by Dissolving 104