# An aqueous solution of a certain organic compound has a density of 1.063 g mL-1 , osmotic pressure of 12.16 atm at 25 °C and a freezing point of 1.03 °C. What is the molar mass of the compound?

### Solution

##### Given: Density of a solution = d = 1.063 g mL-1Osmotic pressure of solution = π = 12.16 atmTemperature = T = 25 °C = 298.15 KFreezing point of solution = Tf = - 1.03 °CTo find: Molar mass of a compoundFormulae: 1. $△{\text{T}}_{\text{f}}={\text{K}}_{\text{f}}\text{m}$  2. π = MRT3. m = $\frac{1000{\text{W}}_{2}}{{\text{M}}_{2}{\text{W}}_{1}}$  Calculation: R = 0.08205 dm3 atm K-1 mol-1$△{\text{T}}_{\text{f}}={\text{T}}_{\text{f}}^{0}-{\text{T}}_{\text{f}}$ = 0 °C - (- 1.03 °C) = 1.03 °C = 1.03 KKf of water = 1.86 K kg mol–1Using formula (i),$△{\text{T}}_{\text{f}}={\text{K}}_{\text{f}}\text{m}$m = $\frac{△{\text{T}}_{\text{f}}}{{\text{K}}_{\text{f}}}=\frac{\text{1.03 K}}{1.86{\text{K kg mol}}^{-1}}$ = 0.554 mol kg-1 = 0.554 mUsing formula (ii),π = MRTM =  = 0.497 mol dm-3 = 0.497 MMass of solvent =  = 0.897 kg = 897 g unitsMass of solution = 1.063 g mL-1 × 1000 mL = 1063 gMass of solute = 1063 g – 897 g = 166 gNow, using formula (iii),m = $\frac{1000{\text{W}}_{2}}{{\text{M}}_{2}{\text{W}}_{1}}$∴ ${\text{M}}_{2}=\frac{1000{\text{W}}_{2}}{\text{m}{\text{W}}_{1}}=\frac{1000{\text{g kg}}^{-1}×166\text{g}}{0.554{\text{mol kg}}^{-1}×897\text{g}}$ = 334 g mol-1The molar mass of the compound is 334 g mol-1.

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