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Theboiling pointof anaqueoussolutionis 100.18.^(@)C. Find thefreezingpointof the solution . (Given: K_(b)= 0.52K kg mol^(-1),K_(f) = 1.86K kgmol^(-1)) |
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Answer» `K_(o) =0.52 K kg MOL^(-1), K_(f) =1.86 K kg mol^(-1)` Boiling point ofwater`= T_(o)= 373 K` `T_(f)= ?` `Delta T_(b)= T_(b) - T_(o) =373 . 18 - 373= 0.18 K` If m is themolalityof thesolutionthen `Delta T_(b)= K_(b ) XX m " and"Delta T_( f) = K_( f) xx m` `:.(DeltaT_(f))/(Delta T_(b))= (K_(f) xx M)/(K_(b) xx M) = (K_(f))/(K_(b))` `:.Delta T_(f) = Delta T _(b) xx .(K_(f))/(K_(b))= 0.18 xx .(1.86)/(0.52) =0.6438 K` Freezingpoint ofwater = `T_(o)= 273 K` `Delta T_(f) = T_(o)- T_(f)` HENCETHE freezingpoint of thesolutionis `T_(f)=T_(o)- Delta T_(f) =273-0.6438= 272. 3562 K` ORFreezingpoint ofsolutionis - 0.6438 `.^(@)C` |
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