InterviewSolution
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InterviewSolution
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Calculate the freezing point of a solution containing 60 g glucose (Molar mass = 180 g `mol^(-1)`) in 250 g of water . (`K_(f)` of water = `1.86 K kg mol^(-1)`) |
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Answer» `DeltaT_(f) = K_(f) xx m ` `T_(f)^(@) - T_(s) = (K_(f) xx W_(B))/(M_(B) xx W_(A) (kg))` `T_(f)^(@)` for water = `0^(@)C = 273 K` `M_(B) (C_(6)H_(12)O_(6)) = 12 xx 6 + 12 xx 1 + 6 xx 16 = 180 g//mol` `273 - T_(s) = (1.86 xx 60 xx 1000)/(180 xx 250)` `273-T_(s) = 2.48` `T_(s) = 273 - 2.48` `T_(s) = 270.52 K` `{:(T_(f)^(@) = "freezing point of solvent") , (T_(s)^(@) = "freezing point of solution." ) , (W_(B) = "given mass of solute"), (M_(B) = "molar mass of solute") , (W_(A) = "mass of solvent") , (K_(f)= "mole depression constant"):}` |
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