Question

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

Answer 1

`Delta T = (1000 xx K xx w)/(m xx W)`
Assuming `5%` weight per cent, we have
`w = 5g, W = 100-5 = 95g`,
`Delta T = 273.15-271 = 2.15, m = 342` (for sugar)
`therefore 2.15 = (1000 xx K xx 5)/(342 xx 95)` …(1)
For glucose `Delta T = (1000 xx K xx 5)/(180 xx 95)` ...(2)
By eqs. (1) and (2), `Delta T = 4.085`
`therefore` Freezing point `= 273.15-4.09`
`=269.06 K`