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 water is 273.15 K.
Open in App
5% by mass of cane sugar means 100 g of solution has 5 g of cane sugar. The molecular weight of cane sugar is 342 g/mol. Number of moles of cane sugar =5g342g/mol=0.0146mol Mass of water =100g−5g=95g=0.095kg (∵1kg=1000g)
The depression in the freezing point ΔTf=273.15K−271K=2.15K ΔTf=Kfm Kf=ΔTfm=2.15K0.154mol/Kg=13.97KKg/mol 5% by mass of glucose means 100 g of
solution has 5 g of glucose. The molecular weight of glucose is 180 g/mol. Number of moles of glucose =5g180g/mol=0.0277mol Mass of water =100g−5g=95g=0.095kg (∵1kg=1000g)
The depression in the freezing point ΔTf=Kfm
The freezing point of glucose solution =273.15K−4.09K=269.07K