Difference of boiling point and freezing point of an aqueous solution of glucose is $ 104 ^ \circ C at 1 bar pressure; then what will be the molality of the solution ? (Kb = 0.513 ^\circ and K+†= 1.86 ^\circ C – kg - mole^{-1}) $
$ T_b - T_f = 104 $ $ T_b ^ o + \triangle T_b - ( T_f ^ \circ - \triangle T_f ) = 104 $ $ \therefore 100 + \triangle T_b - (0 - \triangle T_f ) = 104 $ $ \therefore \triangle T_b + \triangle T_f = 4 $ $ For glucose i=1 \therefore mKb + mKf = 4 \therefore m = { 4 \over (Kb+Kf) } = { 4 \over (0.513 + 1.86 ) } \therefore m = 1.68 $