What mass of helium in grams is required to fill 15.0 L balloon to a pressure of 1.1 atm at 25°c?

Question

asked Sep 27, 2024 33.6k views

1 Answer

Fatemeh Rostami · Answer 1 · 2024-10-01T08:49:44+0000

Answer:

2.77g

Step-by-step explanation:

According to ideal gas law PV=NRT

T= 25+253.15 =298.15K

N= (1.1atm*15L)/(.0821L.atm/mol.K)*(298.15K) = 0.693mol

Molar mass for Helium is 4.0026 g/mol

Mass of helium = N*molar mass = 0.693*4.0026 = 2.77g