A 10% solution (by mass) pf sicrose in water has freezing point of 269.15 K. Calculate freezing point of 10% glucose in water, if freezing point of pu
A 10% solution (by mass) pf sicrose 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)`).
1 Answers
`DeltaT_(f)=(W_(B)xxK_(f))/(M_(B)xxW_(A))`
`Deltat_(f_(sucrose))=(273.15 K-269.15 K)=4.0 K`,
`M_(sucrose)=342g mol^(-1), M_(glucose)=180 g mol^(-1)`
According to availabel date , `W_(B) "and "W_(A)` are same for both the solutes ,` K_(f)` is also same.
` (DeltaT_(f_("scurose")))/(DeltaT_(f_("scurose")))=("Molar mass of glucose")/("Molar mass of glucose")=((180 g mol^(-1)))/((342 g mol^(-1)))`
`DeltaT_(f_("glucose"))=(4.0K)xx((432g mol^(-1)))/((180g mol^(-1)))=7.6 K`
Thus, freezing of glucose solutin= (273.15K-7.60K)=265.55 K.