A sample of gas at 310 Kelvin and 1.2 atmospheres has a density of 2.78 g/L. What is the molar mass of this gas?

1 answer:

5 0

Ideal gas law PV = nRT or PM = dRt ( M is molecular mass of gas ) and d is gas density
where m/ M ( m is mass of gas sample )

m = dRT / P ( 2.78 g/L ) ( 0.082 L atm/ K mole )( 310K ) / 1.2 atm =
58.9 g/mol

gas must be butane ( C4H10) has molecular mass = 58.9 g/mol

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