Ou have been given a 200.00 ml canister containing an unidentified gas. after accounting for the mass of the container, you determine the mass of the actual gas is 0.376 g. the pressure gauge on the cannister reads 874 mmhg, and the temperature gauge reads 298.15 k. what is the identity of the gas? (r= 0.0821 l atm/ mol k)

Ar (argon)

Explanation:

1) Data:

V = 200.00 ml = 0.20000 liter

m = 0.376 g

p = 874 mmHg = 1.15 atm

T = 298.15 K

R = 0.0821 atm×liter / mol×K

2) Formulas:

i) pV = nRT

ii) molar mass = mass in grams / number of moles

3) Solution

i) pV = nRT ⇒ n = pV / RT= 1.15 atm × 0.20000 liter / (0.0821 atm×liter/K×mol × 298K)

n = 0.0094 mol

ii) molar mass = 0.376 g / 0.0094 mol = 39.99 g/mol

Now you have to compare with the molar masses of your set of choices.

The molar mass of argon is 39.95 g/mol, so that is the right choice.

2b2t

explanation:

2b2t