`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`