A 10% solution (by mass) of sucrose in water has freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water, if 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)` ]

Explanation

Molality of Glucose = `10/180 xx 1000/90`
Molality of Sucrose ` = 10/342 xx 1000/90`
`DeltaT_f` (Sucrose) = 273.15 - 269.15 = 4 K
Applying the relation : `Delta T_f = K_f xx ` molality
`Delta T_f` (Glucose) = `K_f` molality (Glucose) ..(1)
`Delta T_f` (Sucrose) = `K_f xx` molality (Sucrose) ...(2)
Dividing (1) by (2), we get
`(Delta T_f("glocose"))/(Delta T_f("sucrose")) = ("Molality (Glucose)")/("Molality (Sucrose) ")`
Substituting the values, we have
`(Delta T_f("glucose"))/(4) = (10 xx 1000)/(180 xx 90) = (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