A 0.1539 molal aqueous solution of cane sugar (mol. mass = 342 g `"mol"^(-1)` ) has a freezing point of 271 K while the freezing point of pure water is 273.15 K. What will be the freezing point of an aqueous solution containing 5 g of glucose (mol. mass = 180 g `"mol"^(-1)` ) per 100 g of solution ?

Explanation

The following data is provided :
`Delta T_f = T_f^0 = 273.15 - 271 = 2.15 K`
Using the following relation and substituting the values, we get
`Delta T_f = K_f xx m`
`2.15 = K_f xx 0.1539`
`K_f = (2.15)/(0.1539)`
` Delta T_f = K_f xx m = (2.15)/(0.1539) xx 5/180 xx 1000/95 = (10750)/(2631.69) = 4.08 K`
Freezing point = 273.15 - 4.08 = 269.07 K.