calculate the freezing point of water (Kf= 1.86 C/m) for an ideal solution of sodium phosphate...

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calculate the freezing point of water (Kf= 1.86 C/m) for anideal solution of sodium phosphate which has a concetration of 3.25m .

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Answer â€“ We are given, molality of Na₃PO₄ = 3.25 m , Kf= 1.86 ^oC/m,

We know pure solvent, means water freezing point = 0.00^oC,

First we need to calculate the freezing point depression and we know formula Â Â

âˆ†Tf = i * Kf * m

We know, I is the van't Hoff factor and for the Na₃PO₄ it is 4

So, âˆ†Tf = 4 * 1.86 ^oC/m * 3.25 m

Â Â Â Â Â Â Â Â Â Â Â Â = 24.18^oC

So, Freezing point of ideal solution of Na₃PO₄ = 0.00^oC - âˆ†Tf

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.00^oC â€“ 24.18^oC

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = â€“ 24.18^oC

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