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Grade 12Mechanics

What is the freezing point of an aqueous solution that was at 104°C?

Profile image of Anshu agrawal
8 Years agoGrade 12
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1 Answer

Profile image of Kuldip Pawar
8 Years ago
use equation for boiling point elevation ∆Tb = Kb x m x i .and this equation for freezing point depression ∆Tf = Kf x m x i .Kb = ebullioscopic constant (aka boiling point elevation constant) for the solvent (water in this case) = 0.512 C/m .Kf = cryoscopic constant (aka freezing point depression constant) for the solvent. = 1.86 C / m for water .m = molality of the solution = moles solute / kg solvent .i = van`t hoff factor = # ions produced in solution / molecule solvent .got that?... the idea here is m and i are constant for the solution. the concentration doesn`t change. neither does the amount of ions in solution. so from the equations above... ∆Tb / Kb = m x i = ∆Tf / Kf .rearranging ∆Tf = ∆Tb x Kf / Kb .since the normal boiling point of water = 100C .∆Tf = (104 C - 100 C) x (1.86 C/m) / (0.512 C/m) = 14.53 C .since the normal freezing point of water = 0C and ∆Tf = 14.53 C, the depressed freezing point = -14.53 C