A sample of an unknown volatile liquid was injected into a Dumas flask (m_flask= 27.0928 g, V_flask= 0.1040 L) and heated until no visible traces of the liquid could be found. The flask and its contents were then rapidly cooled and reweighed (m_flask₊_vapor= 27.4593 g). The atmospheric pressure and temperature during the experiment were 0.976 atm and 18.0 °C, respectively. The unknown volatile liquid was . A) C₇H₁₆ B) C₆H₆ C) C₆H₁₂ D) C₇H₁₄ E) C₆H₁₄

Question

Quizplus · Accepted Answer

The mass of the vapor is 27.4593 g - 27.0928 g = 0.3665 g. Using the ideal gas law, PV = nRT, we can calculate the molar mass. P = 0.976 atm, V = 0.1040 L, T = 18.0 °C = 291.15 K, and R = 0.0821 L·atm/mol·K. Solving for n gives n = PV/RT = (0.976 * 0.1040) / (0.0821 * 291.15) = 0.00416 mol. Molar mass = mass/n = 0.3665 g / 0.00416 mol = 88.0 g/mol, which matches the molar mass of C₆H₁₄.

A Sample of an Unknown Volatile Liquid Was Injected into a Dumas