Answer to Question #165170 in General Chemistry for Eli

Question #165170

What is the boiling point of the solution formed by dissolving .25 mole of a nonvolatile, nonelectrolyte in 250g of water at 1 atm


1
Expert's answer
2021-02-22T05:54:41-0500
"\\Delta\\ T b=Tb(solution)-Tb(pure solvent)"

"\\Delta\\ T b=E*Cm"

E - ebullioscopic constant (0.52 for water)


"Cm=m(solute)*1000\/(M(solute)*m(solvent))=n*1000\/m(solvent)"

"\\Delta Tb=0.52*0.25*1000\/250=0.52\u00b0C"

"Tb(solution)=100+0.52=100.52 \u00b0C"


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