How many moles of gas are present in 57.20 L of argon at a pressure of 1 atm and a temperature of 0°C?

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20 POINTS

n= 2.55 moles

Explanation:

Using the formula of ideal gas law:

PV = nRT

  nRT=PV

n= PV/RT

  n= number of moles

  R= Avogadro constant = 0.0821

  T= Temperature in K => ºC + 273.15 K

  V= volume in L

  P= pressure in atm

n= (1 atm)(57.20 L) / (0.0821)(237.15 K)

n= 2.50 moles