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

Question

Is the 93.67 g/mol the correct answer?

Answer 1

Yes and No. By the Conventional Method of Measurement of Gasoline in a Liquid State, which is 14.7 Parts of Air to 1 Part of Fuel, Yes. But, in a Vaporized State, 100 Parts of Air to 1 Part of Fuel, the "Molar Mass" is Increased many Times Over.My Avatar is Proof of that. Vaporized Gasoline safely Converted to a super-rich Propane. I've Ignited it to Prove that it actually Works.Perfected, this has the Potential to enable even the Largest SUV to get 50 + MPG, and Emit 10 Times less Polluting Exhaust Emissions. And, it's illegal on any Vehicle in the USA from 1996 to the Present, thanks to the EPA-OBD II Law that requires all such Vehicles to Run at 14.7 Parts of Air to 1 Part of Fuel.Want More ? Go to http://www.fuelvapors.com and http://energy21.freeservers.com/bookrep.html Now, just what is "The Correct Answer"?

Answer 2

The molar mass of the gas is constant at any temperature, so the 125 degrees C is extraneous information, especially since you don't know the pressure or the volume at that temperature. STP is 1 atm at 0 C, or 273 K.

PV = nRT
PV = gRT/MM
MM = gRT/PV = 2.15g * 0.08206 *273 K/ (1 atm * 0.750 L)
MM = 64.22 g/mol

Answer 3

If a fifteen.80 3 g pattern of a gasoline occupies 10.0 L at STP, what's the molar mass of the gasoline at one hundred twenty five stages Celsius? --------------------------------------... 15.80 3 g/10 L = x/22.4 L (15.80 3 g x 22.4 liters/mole)/10 liters = x = 35.6 grams/mole. x = molar mass, which does no longer replace with temperature.