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An aqueous solution freezes at `-0.186^(@)C (K_(f)=1.86^(@)` ,`K_(b)=0.512^(@)`. What is the elevation in boiling point?A. `0.0512^(@)C`B. `100.0512^(@)C`C. `-0.0512^(@)C`D. None of these |
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Answer» Correct Answer - A Elevation in boilling point, `DeltaT_(b)=K_(b)xxm` Depression in freezing point `DeltaT_(f)=K_(f)xxm` `(DeltaT_(b))/(DeltaT_(f))=(K_(b))/(K_(f))and(DeltaT_(b))/(0.186)=(0.512)/(186)` `DeltaT_(b)=0.0512^(@)C` |
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