If 125 mg of Ar(g) is added to a 505-mL sample of Ar(g) at STP, what volume will the sample occupy when the conditions of STP are restored

The new volume will be 575 mL

Explanation:

Step 1: Data given

Mass of Ar = 125 mg = 0.125 grams

Volume = 505 mL = 0.505 L

Molar mass Ar = 39.95 g/mol

Step 2 : Calculate moles of Ar to start

n = PV/RT

⇒ with n = initial number of moles Ar

⇒ with p = pressure = 1.0 atm

⇒ with R = the gas constant = 0.08206 L* atm/mol *K

⇒ with V = the volume = 0.505 L

⇒ with T = the temperature = 273.15 K

n = 0.505 L * 1 atm / (0.08206 * 273.15 deg K) =0.02253 moles

Step 3: Calculate add moles of Ar

Moles Ar = mass Ar / molar mass Ar

Moles Ar = 0.125 g / 39.95 g/mol = 0.00313 moles

Step 4: Calculate total moles Ar

Total moles Ar = 0.02253 + 0.00313 = 0.02566 moles

Step 5: Calculate new volume

V = nRT/P

V = 0.02566 * 0.08206 * 273.15 / 1 atm

V = 0.575 L = 575 mL

The new volume will be 575 mL

i think that the answer is c

the awnser is group 12