1 mol occupies 22.4 L at STP. Solve for n, then n = grams/molar mass. You know molar mass and n, solve for grams.
OR you may use PV = nRT to solve for n.
A sample of argon gas at STP occupies 56.2 liters. Find the number of moles of argon and the mass in the sample
3 answers
PV = nRT
n = PV/RT @ STP T= 0+273k; P = 1atm R = 0.821 L- atm/mol-K
n = 1atm(56.2L)/0.821 L- atm/mol-K(273k)
n = 2.50 mole
2.50 moles of Ar =39.984g of Ar/1mole of Ar
n = PV/RT @ STP T= 0+273k; P = 1atm R = 0.821 L- atm/mol-K
n = 1atm(56.2L)/0.821 L- atm/mol-K(273k)
n = 2.50 mole
2.50 moles of Ar =39.984g of Ar/1mole of Ar
Solution:
1) Rearrange PV = nRT to this:
n = PV / RT
2) Substitute:
n = [ (1.00 atm) (56.2 L) ] / [ (0.08206 L atm mol¯1 K¯1) (273.0 K) ]
n = 2.50866 mol (I'll keep a few guard digits)
3) Multiply the moles by the atomic weight of Ar to get the grams:
2.50866 mol times 39.948 g/mol = 100. g (to three sig figs)
1) Rearrange PV = nRT to this:
n = PV / RT
2) Substitute:
n = [ (1.00 atm) (56.2 L) ] / [ (0.08206 L atm mol¯1 K¯1) (273.0 K) ]
n = 2.50866 mol (I'll keep a few guard digits)
3) Multiply the moles by the atomic weight of Ar to get the grams:
2.50866 mol times 39.948 g/mol = 100. g (to three sig figs)
1 answer