1)
mass of KBr dissolved = 27. g
moles of KBr = 27.7 / 119 = 0.227 mol
molality = moles / mass of solvent in kg
= 0.227 / 0.300
molality = 0.756 m
delta Tf = i x Kf x m
6.60 - 4.90 = 2 x Kf x 0.756
Kf = 1.1 oC kg/mol
The normal freezing point of a certain liquid X is 6.60 °C, but when 27. g...
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