0.374 g of an unknown volatile liquid condenses into a 186 mL flask at .708 atm pressure and at 213K. What is the molecular weight of the unknown volatile liquid?
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Hi Huey!
This question requires you to use the ideal gas equation PV=nRT where P represents pressure, V represents volume in liters, n represents moles, R represents the gas constant (most likely 0.08206L atm/moles K but ultimately depends on the units of pressure), and T represents the temperature in kelvin (K=C+273).
Knowing this equation, plug in all the variables you know are given in the problem: (0.708atm)(0.186L)=n(0.08206L atm/moles k)(213K). Notice that I did two things here: I converted mL to L and I also inserted the value of the gas constant. Don't worry about this value because it is always the same so you can either memorize it for future problems or look it up. Now we have one missing variable, n, and that is the amount of moles of the volatile liquid. Use algebra to solve for n and you should get about 7.53x10^-3 moles of the volatile liquid. Now remember that we were given the mass because now to find the molecular weight, we just divide grams by moles: (0.374g)/(7.53x10^-3 moles) = about 49.6 g/mole. So the molecular weight of the volatile liquid is 49.6 grams per mole.
I hope this helped and feel free to ask any more questions if this was confusing :)
This question requires you to use the ideal gas equation PV=nRT where P represents pressure, V represents volume in liters, n represents moles, R represents the gas constant (most likely 0.08206L atm/moles K but ultimately depends on the units of pressure), and T represents the temperature in kelvin (K=C+273).
Knowing this equation, plug in all the variables you know are given in the problem: (0.708atm)(0.186L)=n(0.08206L atm/moles k)(213K). Notice that I did two things here: I converted mL to L and I also inserted the value of the gas constant. Don't worry about this value because it is always the same so you can either memorize it for future problems or look it up. Now we have one missing variable, n, and that is the amount of moles of the volatile liquid. Use algebra to solve for n and you should get about 7.53x10^-3 moles of the volatile liquid. Now remember that we were given the mass because now to find the molecular weight, we just divide grams by moles: (0.374g)/(7.53x10^-3 moles) = about 49.6 g/mole. So the molecular weight of the volatile liquid is 49.6 grams per mole.
I hope this helped and feel free to ask any more questions if this was confusing :)
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The ideal gas law (PV = nRT) can be rearranged to give
PM = dRT
P = gas pressure in atm = 0.708
M = molecular weight of gas in g/mole = ?
d = gas density in g/L = 0.374 g / 0.186 L = 2.01
R = gas constant = 0.0821 L atm / K mole
T = Kelvin temperature = 213
M = dRT / P = (2.01)(0.0821)(213) / (0.708) = 49.6 g/mole
PM = dRT
P = gas pressure in atm = 0.708
M = molecular weight of gas in g/mole = ?
d = gas density in g/L = 0.374 g / 0.186 L = 2.01
R = gas constant = 0.0821 L atm / K mole
T = Kelvin temperature = 213
M = dRT / P = (2.01)(0.0821)(213) / (0.708) = 49.6 g/mole