An aqueus solution freezes at 272.07 K while pure water freezes at 273 K. Determine the molality and boiling point of the solution. Given `K_(f)=1.86 K//m, K_(b)=0.512 K//m.`

Answer

`0,5 m 373.256` K

Explanation

`Delta T_(f) = 273 - 272.07 = 0.93^(@)`
`m = (Delta T_(f))/(k_(f)) = (0.93)/(1.86) = 0.5 m`
Boiling point of solution ` = 373+0.256 = 373.256 K` .