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OpenStudy (anonymous):

The boiling point of an aqueous solution is 101.02 °C. What is the freezing point?

OpenStudy (anonymous):

\[\Delta T=i*Kb *m\] Here since nothing is given i can be safely assumed to be equal to1 . So 1.02=Kb*m. Kb for water is 0.512 so m=1.02/0.512=1.8545 Now substitue in freezing point equation. Kf=1.86 T=1.86*1.8545=3.5. Hence freezing poimt =-3.5 C

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