The problem states that a solution of AlCl3 in 50.0g of waterfreezes at -1.11C.
ID: 677658 • Letter: T
Question
The problem states that a solution of AlCl3 in 50.0g of waterfreezes at -1.11C. Does the molar mas determined from thisfreezing point agree with that calculated from the formula?Why? I think im supposed use the formula delta T of freezing point=the constant freezing point x the molality. I put -1.11C asdelta T of freezing point but i dont know what to do fromthere. The problem states that a solution of AlCl3 in 50.0g of waterfreezes at -1.11C. Does the molar mas determined from thisfreezing point agree with that calculated from the formula?Why? I think im supposed use the formula delta T of freezing point=the constant freezing point x the molality. I put -1.11C asdelta T of freezing point but i dont know what to do fromthere.Explanation / Answer
Formula : Tf = iKf *m Where Tf is the depression in freezing point i is the no.of ions Kfis the molal depression in freezing point constant m is the molality Tf = Tpure -Tsoln For water Tpure = 00C Kf = 1.82 0C / m Data : For AlCl3 , i = 4 as it is forming 4 ions ondissociation Mass ofsolute = 1 g Mass ofsolvent = 50.0 g = 0.05 kg Upon substituting in theabove formula, 1.11 0C = 4 * 1.82 0C / m *m m = 1.11 0C / 4* 1.82 0C / m = 0.152 m But m = mass in g / molar mass * mass ofsolvent molar mass = mass in g / molality * mass ofsolvent = 1 g / 0.152 m * 0.05 kg = 131.57 g / mol It is not exactly matching . There is asmall variation between the calculated value and molar mass of thecompound.Related Questions
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