An aqueous solution of a non electrolyte is having an osmotic pressure of 2 atmosphere at 300 K. Calculate its freezing point. (Given K_f = 1.86 Kg mol^-1, R = 0.821 litre-atm K^-1 mol^-1)

−0.151^oC

$For\; a\; non-electrolyte$

$\backslash [\backslash pi\; =\; MRT\backslash ]$

∵π=2 atm, R=0.0821 L atm K⁻¹mol⁻¹, T=300K

$\backslash (\backslash text\{Therefore,\; \}M\; =\; \backslash frac\{\backslash pi\}\{RT\}\backslash )$

$\backslash (M\; =\; \backslash frac\{2\}\{0.0821\; \times \; 300\}\; =\; 0.0812\backslash )$

$For\; a\; very\; dilute\; solution,\; Molality\; and\; molarity\; are\; same$

$\backslash (\backslash Delta\; T\_f\; =\; K\_f\; m\backslash )$

$\backslash (\backslash Delta\; T\_f\; =\; 1.86\; \times \; 0.0812\; =0.151\backslash )$

$\backslash (\backslash text\{Freezing\; point\}\; =\; T\_o\; -\; \backslash Delta\; T\backslash )$

                                 \(0 - 0.151\)

$\backslash (=\; 0.151^oC\backslash )$