1 litre gas weights 2gat 300k ,1atm if pressure is made at what temperature will 1 litre of same gas weights 1g
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PV = nRT and solve for n (moles)
n = PV/RT = (1atm)(8.2L)/(0.0821 L-atm/K-mol)(300K)
n = 0.333 moles which is represented by 9.0 g
Molar mass = 9.0 g/0.33 moles = 27.3 g/mole
n = PV/RT = (1atm)(8.2L)/(0.0821 L-atm/K-mol)(300K)
n = 0.333 moles which is represented by 9.0 g
Molar mass = 9.0 g/0.33 moles = 27.3 g/mole
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