What mass of argon occupies 4.3 l at 70 kpa and 20 c?

Let us assume that Argon gas is acting Ideally,' Then, According to Ideal Gas Equation,

                                        P V  =  n R T    (1)
Also,
          moles  =  n  =  Mass / M.mass  =  m / M

Substituting value of n in equation 1,

                                        P V  =  (m/M) R T

Solving for mass 'm',

                                        m  =  P V M / R T    (2)
Data Given;
                   P  =  70 kPa  =  0.69 atm

                   V  =  4.3 L

                   T  =  20 °C + 273  =  293 K

                   R  =  0.0821 atm.L.mol⁻¹.K⁻¹

                   M  =  39.94 g.mol⁻¹

Now, putting values in equation 2,

m  =  (0.69 atm × 4.3 L × 39.94 g.mol⁻¹) ÷ (0.0821 atm.L.mol⁻¹.K⁻¹ × 293 K)

m  =  4.89 g of Argon

I'm not sure but i will try to find out

Sorry i dont understand the question can you rephase it?