A 5% solution (by mass) of cane suger in water has freezing point of 271K. Calculate the freezing point of 5% glucose in water if freezing point of water is 273.15 K.

Explanation

Given 5% solution of cane sugar
w = 5 gms, W = 95 gms
m = 342, `DeltaT_("f")=273.15-271=2.15`
`DeltaT_("f")=(K_("f")xxw)/(m xx W)`
`2.15=("K"_("f")xx5)/(342xx95)" …(1)`
5% solution of glucose
w = 5 gms, W = 95 gms ltbrrgt m = 180, `DeltaT_("f")=` ?
`DeltaT_("f")=(K_("f")xx5)/(180xx95)" "...(2)`
Dividing equation (2) by Equation (1)
`(DeltaT_("f"))/(2.15)=(K_("f")xx5)/(180xx95)xx(342xx95)/(K_("f")xx5)`
`DeltaT_("f")=(342xx2.15)/(180)=4.085"K"`
`:.` The freezing point temperature for `5%` glucose solution `=273.15-4.085=269.07" K"`