Question

A mixture of hydrogen and helium gases is maintained in a 9.75 L flask at a pressure of 2.88 atm and a temperature of 76 °C. If the gas mixture contains 0.492 grams of hydrogen, the number of grams of helium in the mixture is ___g.

Answer 1

According to ideal gas law PV = nRT p is the pressure V is the volume n is number of moles R = 0.0821 L. atm. mol.kt t is the temperature in Kelvin. mass molar mass

Pue = nRT PH = eles 0.492 mod X0.0821 L-atmimet". IX 16+273)* 2 X 9.752 PH₂ = 0.722 93 470769 atm 2.88 - 0-72293470769 atm PHe = 2.15706529231 atm PH, XV He= RT lesso (2-1570652923iatr) x (9.75L) He = (0-089 1 Lath. mad K?) (16+273 )* He = 0.734 mod mass= nHQ X molar mass of He mass = 0.734 Mod X 49/mal = 2.9369.