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A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of a 5% glucose in water if freezing point of pure wateris 273.15 K. |
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Answer» SOLUTION :`5%` solution by mass means 5 G of solute is present in 100 g of solution `therefore `Mass of solvent (Water) = 95 g `"Molality of sugar solution "=(5)/(342)xx(1000)/(95)=0.154""("mol. Mass of sugar "C_(12)H_(22)O_(11)=342)` `DeltaT_(f)" for sugar solution "=273.15-271=2.15^(@), DeltaT_(f)=K_(f)xxm therefore K_(f)=(2.15)/(0.154)` `"Molality of glucose solution "=(5)/(180)xx(1000)/(95)=0.292""("mol. mass of glucose "C_(6)H_(12)O_(6)=180)` `therefore""DeltaT_(f)" (Glucose)" =K_(f)xxm=(2.15)/(0.154)xx0.292=4.08` `therefore"FREEZING point of glucose solution "=273.15-4.08="269.07 K."` |
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