A 10% solution (by mass) of sucrose in water has a freezing point of 269.15 K. Calculatethe freezing point of 10% glucose in water if the freezing point of pure water is 273.15 K.[Given : Molar mass of sucrose = 342 g "mol"^(-1), Molar mass of glucose = 180 g "mol"^(-1) ]

A 10% solution (by mass) of sucrose in water has a freezing point of 2

1.	A 10% solution (by mass) of sucrose in water has a freezing point of 269.15 K. Calculatethe freezing point of 10% glucose in water if the freezing point of pure water is 273.15 K.[Given : Molar mass of sucrose = 342 g "mol"^(-1), Molar mass of glucose = 180 g "mol"^(-1) ]
Answer» Solution : MOLALITY of Glucose = `10/180 xx 1000/90` Molality of Sucrose ` = 10/342 xx 1000/90` ` Delta T_f`(Sucrose) = 273.15 - 269.15 = 4 K Applying the relation : `Delta T_f = K_f xx ` molality `Delta T_f`(Glucose) ` = K_f xx `molality (Glucose)...(1) ` Delta T_f` (Sucrose) ` = K_f xx `molality (Sucrose)..(2) Dividing (1) by (2), we GET ` (Delta T_f ("Glucose"))/(Delta T_f ("Sucrose") ) = ("Molality (Glucose)")/("Molality (Sucrose)")` Substituting the values, we have `(Delta T_f ("Glucose"))/(4) = (10 xx 1000)/(180 xx 90)xx(342 xx 90)/(10 xx 1000)= 342/180` ` Delta T_f ("Glucose") = 342/180 xx 4 = 7.6` Freezing POINT of Glucose solution = 273.15 - 7.6 = 265.55 K.