1.

1.8 g of glucose is dissolved in 100 g of water in a beaker. At what temperature will the solution boil if the pressure is 1.013 bar ? Given that the boiling point of pure water at 1.013 ba is 373.15 K and K_(b) for is 0.052K kg mol^(-1).

Answer»


Solution :`M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A)) or DeltaT_(b)=(K_(b)xxW_(b))/(M_(B)xxW_(A))`
`W_(B)=1.8g, W_(A)=100g=0.1" Kg", M_(B)=180" g mol"^(-1), K_(b)=0.052" K kg mol"^(-1)`
`DeltaT_(b)=((0.052" K kg mol"^(-1))xx(1.8 g))/((180" g mol"^(-1))xx(0.1" kg"))=0.005 K`
Boiling POINT of pure water = 373.15 K
Boiling point of solution = (373.15+0.005)K
=373.155 K


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