Calculate the freezing point of an aqueous soltuion of non-electrolyte having an osmotic pressure `2.0 atm` at `300 K`. (`K'_(f) = 1.86 K mol^(-1) kg` and `S = 0.0821` litre atm `K^(-1) mol^(-1)`)

Explanation

We know that,
P = CRT
or `C=(P)/(RT)=(2)/(0.0821xx300)"mol lit"^(-1)`
In dilute solution, the density of water can be taken as `1.0g cm^(-3)`.
Hence, `(2)/(0.0821xx300)"mol lit"^(-1)=(2)/(0.081xx300)"mol kg"^(-1)`
Let `Delta T` be the depression in freezing point
`DeltaT = K_(f)xx` molality
`= 1.86xx(2)/(0.821xx300)=0.151K`
Freezing point `=(273 -0.151) = 272.49 K or - 0.151^(@)C`.