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`H_(2)O (1) hArr H_(2) O(g)` at `373K,DeltaH^(@)=8.31kcal mol^(-1)` Thus, boiling point of 0.1 molal sucrose solution isA. `373.52K`B. `373.052K`C. `373.06K`D. `374.52K` |
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Answer» Correct Answer - C `H_(2)O(l)` changes to steam `H_(2)O (g)`at`373K`.Thus , this represent latent heat of vaporization .`K_(b)`(molal elevaion constant)is related to `DeltaH^(@)` and boiling point by equation, `K_(b)=(RT_(0)^(2))/(1000cancelOH^(@)` Here,`DeltaH^(@)` is in energy unit per gram of solvent. `=(8.31)/(18)kcal^(-1)` `:.K_(b)=(0.002xx(373)^(2))/(1000xx((8.31)/(18)))` `=(278.25)/(461.66)=0.60^(0)mol^(-1)kg` `DeltaT`(surrose solution = molarity `xx K_(b)=0.1xx0.60=0.06^(0)` `:.` Boiling point of solution `= 373+0.06^(@)=373.06K` |
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