A 10% solution (by mass) of sucrose in water has freezing point of 269.15 K. Calculate freezing point of 10% glucose in water, if freezing point of pure is 273.15 K (Given molar mass of sucrose= 342 g `mol^(-1)`, Molar mass of glucose =180 g `mol^(-1)`).

Explanation

`10%` solution in water by mass menas 10 g of sucrose is present in 100 g of the solution, i.e.,
Solute (sucrose) = 10 g, solvent (water) = 90 g
`M_(2)=(1000K_(f)w_(2))/(w_(1)xxDeltaT_(f))`
`"or "K_(f)=(M_(2)xxw_(1)xxDeltaT_(f))/(1000xxw_(2))=(342xx90xx(273.15-269.15))/(1000xx10)="12.312 K kg mol"^(-1)`
`10%` of glucose solution in water means `w_(2)=10, w_(1)=90g`
`DeltaT_(f)=(1000K_(f)w_(2))/(w_(1)M_(2))=(1000xx12.312xx10)/(90xx180)=7.6^(@)`
`therefore" Freezing point of solution "= 273.15 - 7.6=265.55 K`