A 5% solution (by mass) of cane sugar in water has freezing point of 271K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.
Answers
In case of cane sugar:ΔTf = (273.15 - 271) K = 2.15 K (freezing point difference).
Molar mass of sugar (C12H22O11) = 12 × 12 + 22 × 1 + 11 × 16 = 342 g mol – 1.5% solution (by mass) of cane sugar in water means 5 g of cane sugar is present in (100 - 5)g = 95 g of water.
Now, number of moles of cane sugar =5/342 mol= 0.0146 mol.
Therefore, molality of the solution =0.0146mol / 0.095kg = 0.1537 kg mol – 1.
Now applying the relation,ΔTf = Kf × mKf = freezing factor ⇒Kf= ΔTf / m⇒ 2.15K / 0.1537 kg mol-1= 13.99 K kg mol-1.
Molar of glucose (C6H12O6) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol - 15% glucose in water means 5 g of glucose is present in (100 - 5) g = 95 g of water.∴
Number of moles of glucose = 5/180 mol= 0.0278 molTherefore, molality of the solution,m =0.0278 mol / 0.095 kg= 0.2926 mol kg – 1.By applying the relation :ΔTf = Kf × m= 13.99 K kg mol - 1 × 0.2926 mol kg - 1= 4.09 K (approximately).
Hence, the freezing point of 5% glucose solution is (273.15 - 4.09) K= 269.06 K…………Answer