A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% solution (by mass) of glucose in water of the freezing point of pure water is 273.15 K. [Molecular masses glucose `C_(6)H_(12)O_(6) = ` 180 amu, cane sugar `C_(12)H_(22)O_(11)=` 342 amu]

Explanation

Molality of sugar solution `=(W_(2) xx 1000)/(M_(2) xx W_(1)) =(5)/(342)xx (1000)/(95)`
`=0.154"kg mol"^(-1) or 0.154 m`
`DeltaT_(f)=T_(f)^(@)-T_(f)=273.15 -271 =2.15K`
`DeltaT_(f)=K_(f) xx m`
`rArr K_(f)=(DeltaT_(f))/(m)=(2.15)/(0.154)`
Molality of glucose solution, `(W_(2) xx 1000)/(M_(2)W_(1)=(5)/(180) xx (1000)/(95)`
`=0.292m`
`:.DeltaT_(f)("glucose")=K_(f) xx m =(2.15)/(01.54)xx 0.292 =4.08`
`:.` Freezing point of glucose solution `= 273.15–4.08`
= 269.07K