1.

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

Answer»

Solution : Molality of SUGAR solution =` 5/342 xx 1000/100 = 0.146 `
`Delta T_f` for sugar solution = 273.15 - 271 = `2.15^@`
USING the following relation and substituting values, we have
Molality of glucose solution = `5/180 xx 1000/100 = 0.278`
Using the relation`Delta T_f xx K_f xx m`and substituting values again, we have
`Delta T_f ` (Glucose) ` = (2.15)/(0.146)xx 0.278 = 4.09^@`
`THEREFORE `Freezing point of glucose solution = 273.15 - 4.09 = 269.06 K.


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