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

A 5% solution (by mass) of cane sugar in water has freezing point of 2

1.	A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the frezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.
Answer» Solution :Here, `Delta T_(f)=(273.15-271)K` = 2.15 K Molar mass of sugar `(C_(12)H_(22)O_(11))` `=12xx12+22xx1+11xx16` `= 342 g MOL^(-1)` 5% solution (by mass) of cane sugar in water MEANS 5 g of cane sugar is present in (100 g - 5)g = 95 g of water. Now, number of cane sugar `= (5)/(342)=0.0146` mol Therefore, molality of the solution, `m=(0.0146 mol)/(0.095 kg)=0.1537 mol kg^(-1)` Applying the RELATION, `Delta T_(f)=K_(f)m` `therefore K_(f)=(Delta T_(f))/(m)` `= (2.15 K)/(0.1537" K kg mol"^(-1))` = 13.99 K kg/mol Molar of glucose `(C_(6)H_(12)O_(6))` `= 6xx12+12xx1+6xx16` `= 180 g mol^(-1)` 5% glucose in water means 5 g glucose is present in (100 - 5) g = 95 g of water. Therefore, Number of moles of glucose `= (5)/(180)` mol = 0.0278 mol Therefore, molality of the solution, `m =(0.0278 mol)/(0.095 kg)` Applying the relation, `Delta T_(f)=K_(f)XX m` = 13.99 K kg `mol^(-1)xx 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.