# A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of a 5% glucose (by mass) in water. The fre

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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 (by mass) in water. The freezing point of pure water is 273.15 K.

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Mass of sugar in 5% (by mass) solution means 5gin 100g of solvent (water)
Molar mass of sugar=342 " g mol"^(-1)
Molality of sugar solution =(5xx1000)/(342xx100)=0.146
therfore DeltaT_(f) for sugar solution =273.15-271=2.15^(@)
Delta T_(f)=K_(f)xxm
Delta T_(f)=K_(f)xx0.146 implies K_(f)=2.15//0.146
Molality of glucose solution
=(5)/(180)xx(1000)/(100)=0.278
(Molar mass of glucose=180 " g mol"^(-1))
Delta T_(f)=K_(f)xxm=(2.15)/(0.146)xx0.278=4.09^(@)
therefore Freezing point of glucose solution
=273.15-4.09=269.06 K.