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InterviewSolution
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A 10% solution (by mass) of sucrose in water has freezing point of 269.15 K. Calculate freezing point of 10% glucose in water, if freezing point of pure is 273.15 K (Given molar mass of sucrose= 342 g `mol^(-1)`, Molar mass of glucose =180 g `mol^(-1)`). |
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Answer» 10% solution (by mass) of sucrose: `{:(w_(B) = 10 g),(m_(B) = 342 g//mol),(w_(S) = 100g"," w_(A) = 100 - 10 = 90g):}` `[(w_(B) = "Mass of solute"),(w_(S) = "Mass of solution"),(w_(A) = "Mass of solvent")]` `Delta T_(f) = k_(f) xx (w_(B))/(m_(B) xx w_(A) (kg))` `T_(A)^(@) -T_(S)^(@) = k_(f) xx (w_(B))/(m_(B) xxw_(A) (kg))` `273.15 - 269.15 = k_(f) xx (10 xx 1000)/(342 xx 90)` `k_(f) = (4 xx 342 xx 90)/(10000) rArr (342 xx 9)/(250)` 10% solution (by mass) of glucose `{:(w_(B) ",",m_(B) = 180 g//mol),(w_(S) = 100g",",w_(A) = 90g):}` `T_(A)^(@) - T_(S) = k_(f) xx (w_(B))/(m_(B) xx w_(A) (kg))` `273.15 - T_(S) = (342 xx 9)/(250) xx (10 xx 1000)/(180 xx 90)` `273.15 - T_(S) = 7.6 k` `T_(S) = 273.15 - 7.6` `T_(S) = 265.55 k` |
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