What is the volume of 10.0 g of argon gas at 157°c and 2.50 kpa pressure?

Data Given:
                  Pressure = P =  2.5 kPa  =  0.02467 atm

                  Temperature = T = 157 °C + 273.15  =  430.15 K

                  Volume = V = ?

                  Mass = m = 10 g

First find out moles for given amount of Argon gas,
As,
                             Moles  =  Mass / M.mass
Putting values,
                             Moles  =  10 g / 39.94 g/mol  =  0.250 moles

Now using Ideal Gas Equation,

                                   P V  =  n R T
Solving for V,
                                   V  =  n R T / P
Putting Values,

         V  =  (0.25 mol × 0.0821 L.atm.mol⁻¹.K⁻¹ × 430.15 K) ÷ 0.02467 atm

                                          Volume =  357.87 L

So what you would do is go through and add 3 teaspoons(tsp) 6 times because of the 6 tablespoons(tbsp). which will give you an answer of 18. another way is to multiply 3x6=18.