A solution is prepared by dissolving 53.5 grams of ammonium chloride, NH₄Cl in 4000. 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) 100.13°C B) 100.26°C C) 100.39°C D) 100.52°C

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Quizplus · Accepted Answer

The boiling point elevation is given by the equation:ΔTb = Kb * molalitywhere Kb is the boiling point elevation constant and molality is the number of moles of solute per kilogram of solvent.In this case, we have:Kb = 0.52°C/mole solute in 1000g of watermolality = (53.5 g NH4Cl / 53.5 g/mol) / (4000 g water / 1000 g/kg) = 0.0125 mol/kgΔTb = 0.52°C/mole solute in 1000g of water * 0.0125 mol/kg = 0.0065°CTherefore, the boiling point of the solution is:100.00°C + 0.0065°C = 100.013°CThe closest answer choice is A) 100.13°C.

A Solution Is Prepared by Dissolving 53