7. how many moles of argon are there in 20.0 l, at 25 degrees celsius and 96.8 kpa?

              0.8133 mol

Data Given:

                 Moles  =  n  =  ??

                 Temperature  =  T  =  25 °C + 273.15  =  298.15 K

                  Pressure  =  P  =  96.8 kPa  =  0.955 atm

                  Volume  =  V  =  20.0 L

Formula Used:

Let's assume that the Argon gas is acting as an Ideal gas, then according to Ideal Gas Equation,

                  P V  =  n R T

where;  R  =  Universal Gas Constant  =  0.082057 atm.L.mol⁻¹.K⁻¹

Solving Equation for n,

                  n  =  P V / R T

Putting Values,

                  n  =  (0.955 atm × 20.0 L) ÷ (0.082057 atm.L.mol⁻¹.K⁻¹ × 298.15 K)

                 n  =  0.8133 mol