Question

What mass of helium, in grams, is required to fill a 6.1-l balloon to a pressure of 1.1 atm at 25 °c?

Answer 1

PV=nRT
(1.1)(6.1)=(n)(0.08201)(298)
6.71=n(0.08201*298)
n=0.2745613769 moles Helium
n(Mm)=mass
0.2745613769(4.00)
1.098245508 grams of Helium