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

`DeltaT_(f)=(K_(f)xxW_(B))/(M_(B)xxW_(A))`
`" For cane sugar solution:"W_(B)=5g, W_(A)=100-5 =95g=0.095 kg,`
`M_(B)=342 g mol^(-1)`
`2.15K=(K_(f)xx(5g))/((342 g mol^(-1))xx(0.095 kg))`
`"For glucose soolution:" W_(B)=5g,W=100-5=95g, =0.095 kg`
`M_(B)=180 g mol^(-1)`
`DeltaT_(f)=(K_(f)xx(5g))/((180g mol^(-1))xx(0.095 kg))`
`"Dividing ean. (ii) by eqn. (i). (DeltaT_(f))/((2.15K))=((K_(f))xx(5g))/((180 g mol^(-1))xx(0.095))xx((342 g mol^(-1))xx(0.095 kg))/((K_(f))xx(5g))`
`DeltaT_(f)=((342g mol^(-1)))/((180 g mol^(-1)))xx2.15 K=4.085 K`
`"Freezing point temperature of glucose solution"=(273.15-4.085) K = 269.07 K`.