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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 water is 273.15 K.

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Solution

## 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 =5 g342 g/mol=0.0146 molMass of water =100 g−5 g=95 g=0.095 kg(∵1 kg=1000 g)Molality m=0.0146 mol0.095 kg=0.154 mol/kgThe depression in the freezing point ΔTf=273.15 K−271 K=2.15 KΔTf=KfmKf=ΔTfm=2.15 K0.154 mol/Kg=13.97 K Kg/mol5 % 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 =5 g180 g/mol=0.0277 molMass of water =100 g−5 g=95 g=0.095 kg(∵1 kg=1000 g)Molality m=0.0277 mol0.095 kg=0.292 mol/kgThe depression in the freezing point ΔTf=KfmΔTf=0.292 mol/13.97 K Kg/mol×0.292 mol/kg=4.09 KThe freezing point of glucose solution =273.15 K−4.09 K=269.07 K  Suggest Corrections  0      Similar questions  Related Videos   Depression in Freezing Point
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