A solution prepared by dissolving 3.27 g of a nonvolatile, covalent solute in 30.0 g of water boils at 100.74 ºC. What is the approximate molar mass of the solute? Kbp H2O = 0.512 ºC/m.

asked by @kinzas1 • about 2 years ago • Chemistry • 5 pts
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1 answer

Hey there. So for this question they are talking about boiling point so you would use the boiling point elevation equation.

Delta Tb = i x Kb x m

The normal boiling point of water is 100 degrees Celsius, but in this question it boils at 100.74 degrees Celsius. So delta Tb would be 0.74 degrees Celsius. Next they say the solute is a nonvolatile, covalent solute so that means it doesn't break up into ions so i = 1. m is just molality, which is moles of solute/kg of solvent.

Delta Tb = i x Kb x m 0.74 = (1) (0.512) (x moles of solute / 0.030 kg water)

solve for the moles of solute to give you 0.043359 moles.

Molar mass is grams / moles so that means 3.27 g / 0.043359 moles = 75.42 g/mol.

Hope that helps.

answered by @jules • about 2 years ago