If a 2.15-g sample of a gas occupies 750. mL at STP, what is the molar mass of the gas at 125°C?

Question

Quizplus · Accepted Answer

Since the gas is at STP, we can use the ideal gas law to find the number of moles of gas: PV = nRT, where P = 1 atm, V = 750 mL = 0.75 L, T = 273 K. Solving for n, we get n = PV/RT = (1 atm)(0.75 L)/(0.0821 L·atm/mol·K)(273 K) = 0.0317 mol.
Next, we can use the formula for molar mass: M = m/n, where m is the mass of the gas. Substituting in m = 2.15 g and n = 0.0317 mol, we get M = 67.9 g/mol. However, this is at 273 K, and we need to find the molar mass at 125°C, which is 398 K. To do this, we can use the formula M1/T1 = M2/T2, where M1 = 67.9 g/mol, T1 = 273 K, T2 = 398 K. Solving for M2, we get M2 = (M1)(T2/T1) = (67.9 g/mol)(398 K/273 K) = 98.8 g/mol, which rounds to 64.2 g/mol. Therefore, the answer is C.

If a 215-G Sample of a Gas Occupies 750