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Question 9.15: What is the boiling point of a solution of 0.75 mol of KBr i...

What is the boiling point of a solution of 0.75 mol of KBr in 1.0 kg of water?
ANALYSIS The boiling point increases 0.51 °C for each mole of solute per kilogram of water. Since KBr is a strong electrolyte, there are 2 mol of ions (K+K^+ and Br−Br^-) for every 1 mol of KBr that dissolves.
BALLPARK ESTIMATE The boiling point will increase about 0.5 °C for every 1 mol of ions in 1 kg of water. Since 0.75 mol of KBr produce 1.5 mol of ions, the boiling point should increase by (1.5 mol ions) × 0.5 °C/mol ions) = 0.75 °C

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ΔTboiling=(0.51 ∘Ckg watermol ions)(2 mol ions1 mol KBr)(0.75 mol KBr1.0 kg water)=0.77 ∘C\Delta T_{boiling} = \left( 0.51  ^\circ C \frac{\cancel{kg  water}}{\cancel{mol  ions}} \right) \left(\frac{ 2  \cancel{ mol  ions}}{1  \cancel{mol  KBr}} \right) \left(\frac{0.75  \cancel{mol  KBr}}{1.0  \cancel{kg  water}} \right) = 0.77  ^\circ C

The normal boiling point of pure water is 100 °C, so the boiling point of the solution increases to 100.77 °C.
BALLPARK CHECK: The 0.77 °C increase is consistent with our estimate of 0.75 °C.

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