A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.


5% solution means 5g of cane sugar is present in 100g of solution

Freezing point of solution = 271k


Freezing point of pure water = 273.15k


Molar mass of cane sugar(C12H22O11) = 12 × 12 + 1 × 22 + 16 × 11 = 342g


Moles of cane sugar = mass/molar mass = 5/342


n = 0.0146mol


Molality of solution = moles of solute/mass of solvent(in kg)


M = 0.0146/0.095


Molality = 0.154M


Depression in freezing point = ΔTf = 273.15-271 = 2.15k


Applying the formula: ΔTf = Kf × M


Where


ΔTf = depression in freezing point


Kf = molal depression constant


M = molality of solution


Kf = 2.15/0.154


Kf = 13.96k kg mol-1


Second condition: mass of glucose = 5g


Molar mass of glucose(C6H12O6) = 12 × 6 + 1 × 12 + 16 × 6 = 180g


Moles of glucose = mass/molar mass


n = 5/180moles


n = 0.0278mol


Molality of solution = moles of solute/mass of solvent(in kg)


molality = 0.0278/0.095


M = 0.2926M


Applying the formula: ΔTf = Kf × M


Where


ΔTf = depression in freezing point


Kf = molal depression constant


M = molality of solution


ΔTf = 13.96 × 0.2926


ΔTf = 4.08k


Freezing point of solution = 273.15 + 4.08 = 277.234k


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