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Solution of sucrose is prepared by dissolving 34.2 g of it in 1000 g of water. Find out the freezing point of the solution, if Kf of water is 1.86 K/kg/mol? (Molecular mass of sucrose is 342 g/mol). |
Answer» ΔTf = kf × m = 1.86 K kg mol-1 × \(\frac{34.2\times1000g\,kg^{-1}}{342g\,mol^{-1}\times\,1000g}\) = 0.186 K Tf = OK – 0.186 K Freezing point of the solution = – 0.186 K |
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