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 __________.

Question

Quizplus · Accepted Answer

The molar mass of the unknown liquid can be calculated using the ideal gas law and the mass difference. The mass of the vapor is 0.3665 g. Using PV = nRT, where P = 0.976 atm, V = 0.1040 L, R = 0.0821 L·atm/mol·K, and T = 291.15 K, we find n = 0.00423 mol. The molar mass is then 0.3665 g / 0.00423 mol = 86.6 g/mol, which is closest to the molar mass of C₆H₁₄ (hexane).

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