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Questions 4-15 · Q12

Q.An aqueous solution of a certain organic compound has a density of 1.063 gmL⁻¹, an osmotic pressure of 12.16 atm at 25⁰C and a freezing point of -1.03⁰C. What is the molar mass of the compound? (334 g/mol)

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Step 1. From osmotic pressure (π=MRT\pi=MRT): M=π/(RT)=12.16/(0.08206×298)=0.4973M = \pi/(RT) = 12.16/(0.08206\times298) = 0.4973 mol/L (25 degC = 298 K).

Step 2. From freezing point depression (ΔTf=Kfm\Delta T_f=K_fm), using the STANDARD value Kf(water)=1.86K_f(\text{water})=1.86 K kg mol-1 (not explicitly restated in this problem, but standard for water and used consistently elsewhere in this chapter, e.g. Problems 2.10-2.14): m=ΔTf/Kf=1.03/1.86=0.5538m = \Delta T_f/K_f = 1.03/1.86 = 0.5538 mol/kg.

Step 3. Relating molarity M and molality m through the solution's density rho (1.063 g/mL = 1.063 kg/L): taking 1 L of solution, mass = 1000 x 1.063 = 1063 g; moles of solute = M = 0.4973 mol, so mass of solute = M x M2 grams, and mass of solvent = 1063 - M.M2 (in g) = (1063-M.M2)/1000 kg. Then m=M(1063−M⋅M2)/1000=1000M1063−M⋅M2m = \dfrac{M}{(1063-M\cdot M_2)/1000} = \dfrac{1000M}{1063-M\cdot M_2}, which rearranges to M2=1000(mρ−M)mMM_2 = \dfrac{1000(m\rho-M)}{mM}.

Step 4. Substituting: mρ=0.5538×1.063=0.5887m\rho = 0.5538\times1.063=0.5887; mρ−M=0.5887−0.4973=0.0914m\rho-M=0.5887-0.4973=0.0914; 1000×0.0914=91.41000\times0.0914=91.4; mM=0.5538×0.4973=0.2754mM=0.5538\times0.4973=0.2754; M2=91.4/0.2754≈332M_2=91.4/0.2754\approx332 g/mol. …

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