Blood freezes at `272.44 K` and a solution of `3.0 g` of urea in `250 g` of water freezes at `272.63 K`. Calculate the osmotic pressure of blood at `300 K`. (Assume density of blood at `300 K` to be `1 g c c^(-1)`)

Explanation

In this question first calculate `K_(f)` of water from urea solution.
`DeltaT_(f)=K_(f)(((W_(B))/(Mw_(B)))/(W_(A))xx1000)`
`rArrK_(f)=(DeltaT_(f))/(((W_(B))/((Mw_(B))/(W_(A)))xx1000))`
`implies K_(f)=(0.37)/((((3)/(60))/(250)xx1000))=1.85`
`[Delta T_(f)=273-272.63=0.37 K]`
Now determine the molality of blood by using
`DeltaT_(f)=K_(f)m`
Now `DeltaT_(f)=273-272.44=0.56^(@)C`
`m=(DeltaT_(f))/(K_(f))=0.56/1.85=0.303`
`rArrMolarity=molality=0.303`
Now using `pi=CRT`
`rArr pi =0.303xx0.082xx300=7.46 atm`