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

Explanation

Molar mass of cane sugar
`C_(12) H _(22) O _(11) = 342 g mol ^(-1)`
Molality of sugar `= (5 xx 100)/( 342 xx 100) = 0.146`
`Delta T_(f)` for sugar solution `=273.15 - 271=2.15^(@)`
`Delta T_(f) = K_(f) xxm`
`K _(f) = (2.15)/(0.146)`
Molality of glucose solution `= (5)/(180) xx (1000)/(100) = 0.278 =4.09^(@)K`
`therefore` Freezing point of glucose solution `=273.15-4.09 =269.06K`