1.

18 g of glucose (C_(6)H_(12)O_(6)) is dissolved in 1 kg of water in a saucepn. At what temperature will water boil under 1.013 bar pressure ? Given K_(b) for water is 0.52 K kg mol^(-1)

Answer»

Solution :`W_(B)=(M_(B)xxDeltaT_(b)xxW_(A))/K_(b) or DeltaT_(b)=(W_(B)xxK_(b))/(M_(B)xxW_(A)`
Mass of solute `(W_(B))`=18g
Mass of WATER `(W_(B))`=1 kg
Molar mass of solute `(M_(B))`= 0.52 K kg `mol^(-1)`
Molal elevation constant `(K_(b))`=0.52 K kg `mol^(-1)`
`DeltaT_(b)=((18g)XX(0.52 K kg mol^(-1)))/((180 g mol^(-1))xx(1 kg))=0.052 K`
At 1.013 bar PRESSURE (atmospheric pressure), BOILING point of water= 37300K
Boiling point of solution= `(373.0+0.052 K)`


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