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

`because pi = 2 atm, S = 0.0821" litre atm "K^(-1) mol^(-1)` and `T = 300 K`
`because piV = nST`
`:. (n)/(V) = (pi)/(ST) = (2)/(0.0821 xx 300)`
`:.` Molarity `((n)/(V)) = 0.0821 mol litre^(-1)`
Since, `Delta T = K'_(f) xx " molality"`
For dilute solutions, molarity = molality
and thus, molality `= 0.0812`
`:. Delta T = 1.86 xx 0.0812 = 0.151`
`:.` Freezing point `= T_(0) - (DeltaT) , (because T_(0) =0^(@)C)`
`= 0-0.151 = - 0.151^(@)C`