What volume would a container be at if it held 21.4g of argon at a temperature of 12.5°c and a pressure of 1.25atm?

You will use the formula PV=nRT
P = pressure
V = volume
n = moles
R = gas constant
T = Temperature

Since 21.4g of Ar (Argon) is given this will be divided by the molar mass of Ar to give you the number of moles. Mass of Ar is 40g so

21.4g/40g = 0.54 moles of Argon

Then you just change Celsius to Kelvin by adding 273 to 12.5 which will be

12.5 + 273 = 285.5 K

(The formula for changing C to K is
X°C + 273 = K)

Now with moles and temperature in K you can plug all your values into PV=nRT to find volume so

(1.25atm)V = (0.54moles)(0.0821)(285.5K)

(0.0821 is the gas constant R when using atm, Liters, moles, and kelvin; this should always be given by your teacher)

Multiply
1.25v= 12.66

Divide 1.25 on both sides
V= 12.66/1.25 —> 10.13L

Your answer should be approximately 10.13L

I think the answer is B for 1. And D for 2.