Asked by lola
what is the freezing point of an aqueous solution that boils at 102.1 C?
Kf=1.86
Kb=0.512
102.1-100.0=2.1
2.1/0.512=4.10
4.10*1.86=7.62 C
7.62 C
Is this correct?
Kf=1.86
Kb=0.512
102.1-100.0=2.1
2.1/0.512=4.10
4.10*1.86=7.62 C
7.62 C
Is this correct?
Answers
Answered by
Courage
Yes
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