45 gram of Ethylene glycol is mixed with 600 gram of water calculate the freezing point of depression and the freezing point of the solution
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Molar mass of ethylene = 62g/mol
no. of moles of ethylene glycol = 45/62 = 0.73 mol
Mass of Water = 600/1000=0.6
Molality of ethylene glycol= 0.73mol/0.60kg=1.2m
Freezing point depression
▲Tf =kf×m = 1.86×1.2=2.2kelvin.
Freezing point =273.15K-2.2K = 270.95K
no. of moles of ethylene glycol = 45/62 = 0.73 mol
Mass of Water = 600/1000=0.6
Molality of ethylene glycol= 0.73mol/0.60kg=1.2m
Freezing point depression
▲Tf =kf×m = 1.86×1.2=2.2kelvin.
Freezing point =273.15K-2.2K = 270.95K
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