There\'s an unknown solution that freezesat -5.01 C. since Tf= kf*i*m, 5.01=1.86

Question

There's an unknown solution that freezesat -5.01 C. since Tf= kf*i*m, 5.01=1.86*2*m m=1.347 i found til here, which m = moles of solute/kg ofsolvent. After this, I need to find the molecular weight of the unknownsolid. Any help :S? Oh one last thing, it says find (rectangle)T, what would bethe difference between t and (rectangle)Tf? What would be T in this case? There's an unknown solution that freezesat -5.01 C. since Tf= kf*i*m, 5.01=1.86*2*m m=1.347 i found til here, which m = moles of solute/kg ofsolvent. After this, I need to find the molecular weight of the unknownsolid. Any help :S? Oh one last thing, it says find (rectangle)T, what would bethe difference between t and (rectangle)Tf? What would be T in this case?

Explanation / Answer

a )Whatever you did upto this is right.After finding themolality you can find the no.of mols . But no.of mols = Mass of solute in g / molar mass of solute ing / mol . To solve this type of problems definitely mass of solutewill be provided. So you can know the molar mass of the unknownsolute. b) Depressoin in freezing point is nothing but the decrease inthe freezing point of a pure solvent on adding solute. That can be represented as Tf = Tpure -Tsoln You have the Tf   and solvent will bementioned. So for that freezing point can be known . Now Tsoln canbe obtained by subtracting Tf from T pure.

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