1.

A 0.1539 molal aqueous solution of cane sugar (mol. mass = 342 g "mol"^(-1) ) has a freezing pointof 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 ?

Answer»

SOLUTION :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.


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