Calculate the freezing point of the solution when 45g of ethylene glycol (C2H6O2) is mixed with 600g of water.
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Molar mass of ethylene glycol = 62 g/mol
Therefore, number of moles of ethylene glycol = 45/62 = 0.73 mol
Mass of water = 600g/1000 = 0.6 kg
Thus, molality of ethylene glycol = 0.73mol/0.60kg = 1.2 m
Therefore freezing point depression,
∆Tf = Kf x m where, Kf is the molal freezing point depression constant.
= 1.86 K kg mol–1 × 1.2 mol kg –1
= 2.2 K
Hence, freezing point of the aqueous solution = 273.15 K – 2.2 K = 270.95 K
that's the answer hope it helps you
Therefore, number of moles of ethylene glycol = 45/62 = 0.73 mol
Mass of water = 600g/1000 = 0.6 kg
Thus, molality of ethylene glycol = 0.73mol/0.60kg = 1.2 m
Therefore freezing point depression,
∆Tf = Kf x m where, Kf is the molal freezing point depression constant.
= 1.86 K kg mol–1 × 1.2 mol kg –1
= 2.2 K
Hence, freezing point of the aqueous solution = 273.15 K – 2.2 K = 270.95 K
that's the answer hope it helps you
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