An organic compound was prepared and purified by chromatography. A conductivity measurement showed the compound to be a nonelectrolyte. Elemental analysis gave an empirical formula of CH₃O. A 0.127 g sample of the compound was dissolved in water and diluted to 100.0 mL, and the osmotic pressure of that solution was determined to 0.492 atm at 20^$\circ$C. Determine the molar mass of the compound. (R = 0.0821 L·atm/K·mol)

Question

Quizplus · Accepted Answer

The osmotic pressure equation is used: π = MRT. Solving for M (molarity), M = π / (RT). Substituting the given values: M = 0.492 atm / (0.0821 L·atm/K·mol × 293 K) = 0.0204 mol/L. The number of moles in 100 mL (0.1 L) is 0.0204 mol/L × 0.1 L = 0.00204 mol. The molar mass is then 0.127 g / 0.00204 mol = 62.25 g/mol, which is closest to 62.1 g/mol.

An Organic Compound Was Prepared and Purified by Chromatography ∘\circ∘ C Determine the Molar Mass of the Compound

An Organic Compound Was Prepared and Purified by Chromatography $\circ$ C Determine the Molar Mass of the Compound