Question

How many grams of Kr are in a 3.91 L cylinder at 33.0 ∘C and 8.86 atm?

Answer 1

we have

P = 8.86 atm

V = 3.91 L

T = (273 + 33) = 306 K

ideal gas equation is

PV = nRT

or

8.86 atm * 3.91 L = n * 0.0821 L-atm / mole-K * 306 K

or

n = 1.38 mole.

mole of Kr gas = 1.38 mole.

and

mass of Kr = mole * molar mass = 1.38 mole * 83.798 g / mole = 115.6 grams (answer)