A 0.334 g sample of an unknown halogen occupies 109 mL at 398 K and 1.41 atm.What is the identity of the halogen? A) Br₂ B) F₂ C) Cl₂ D) I₂ E) Ge

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To identify the halogen, we can use the ideal gas law, PV = nRT, where P is the pressure in atm, V is the volume in liters, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin. First, convert the volume to liters: 109 mL = 0.109 L. Then, solve for n (moles of gas): n = PV / RT = (1.41 atm * 0.109 L) / (0.0821 L·atm/(mol·K) * 398 K) = 0.00454 mol. The molar mass (M) of the gas can be found by dividing the mass of the sample by the number of moles: M = 0.334 g / 0.00454 mol = 73.57 g/mol. This molar mass is closest to that of chlorine (Cl), which is approximately 35.5 g/mol for one atom, but since Cl exists as a diatomic molecule (Cl2) in nature, its molar mass is approximately 71 g/mol, making chlorine the correct answer.

A 0334 G Sample of an Unknown Halogen Occupies 109 ML