Question

Define cryoscopic constant and write formula for determining the molar mass of a solute using freezing point depression method​

Answer 1

Answer:

In thermodynamics, the cryoscopic constant, Kf, relates molality to freezing point depression (which is a colligative property). It is the ratio of the latter to the former:

{\displaystyle \Delta T_{f}=i\cdot K_{f}\cdot b}{\displaystyle \Delta T_{f}=i\cdot K_{f}\cdot b}

{\displaystyle i}i is the van 't Hoff factor, the number of particles the solute splits into or forms when dissolved.

{\displaystyle b}b is the molality of the solution.

Through cryoscopy, a known constant can be used to calculate an unknown molar mass. The term "cryoscopy" comes from Greek and means "freezing measurement." Freezing point depression is a colligative property, so {\displaystyle \Delta T}{\displaystyle \Delta T} depends only on the number of solute particles dissolved, not the nature of those particles. Cryoscopy is related to ebullioscopy, which determines the same value from the ebullioscopic constant (of boiling point elevation).

The value of {\displaystyle K_{f}}{\displaystyle K_{f}}, which depends on the nature of the solvent can be found out by the following equation:

{\displaystyle K_{f}={\frac {R\cdot M\cdot T_{f}^{2}}{\Delta _{fus}H}}}{\displaystyle K_{f}={\frac {R\cdot M\cdot T_{f}^{2}}{\Delta _{fus}H}}}

{\displaystyle R}R is the Universal Gas Constant

{\displaystyle M}M is the Molar mass of the solvent in {\displaystyle kg\ mol^{-1}}{\displaystyle kg\ mol^{-1}}

{\displaystyle T_{f}}{\displaystyle T_{f}} is the freezing point of the pure solvent in kelvin

{\displaystyle \Delta _{fus}H}{\displaystyle \Delta _{fus}H} represents the molar enthalpy of fusion of the solvent in Joules per mole.