If 125mg of Ar is added to a 505ml sample of Ar at STP, what volume will the sample occupy when the conditions

Question

STP are restored??

help please

i am so confused

Answer 1

n = PV/RT
Moles of Ar to start-
n = 0.505 L x 1 atm / (0.08206 x 273.15 deg K) =0.02253 moles
Added moles of Ar-
125 mg Ar = 0.125 g / 39.944 g/mol = 0.00313 moles
Total moles Ar = 0.02253 + 0.00313 = 0.02566 moles
New volume-
V = nRT/P
V = 0.02566 x 0.08206 x 273.15 / 1 atm
V = 0.575 L = 575 mL

Answer 2

First, from 125mg of argon we need to find the volume:

Argon weighs 40g/mole and it is known that 1 mole of any gas at STP will occupy 22.4 liters so we just need to determine what volume 0.125g occupies:

22.4 L x (0.125g/40g)= 0.0700liter or 70 ml

So total volume would be 505 + 70ml = 575ml

Answer 3

i'm hoping it's 505mg
i'd use V1n2=V2n1 since T & P are constant
V1=1L
V2=?
n1=(0.505/39.95)
n2=(.630/39.95)
V1n2/n1=V2
i get 1.25 L