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A solution of urea in water boills at 100^(@)C. Calculate the freezing point of the same solution. Molal constant K_(f) and K_(b) are 1.86 and 0.521 k m^(-1) repectivly. |
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Answer» `(DeltaT_(f))/(DeltaT_(b))=K_(f)/K_(b)` `DeltaT_(b)=100.18^(@)C-100.0=0.18^(@)C=0.18 K` `K_(f)=1.86" K m"^(-1), K_(b)=0.512" K m"^(-1)` `DeltaT_(f)=K_(f)/K_(b)xxDeltaT_(b)=((1.86"Km"^(-1)))/((0.512"Km"^(-1)))xx0.18K=0.654K=0.654^(@)C` `therefore "Freezing point of soluion"=0^(@)C-0.654^(@)C=-0.654^(@)C.` |
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