96.0 g. of a gas occupies 48.0 L at 700.0 mm Hg and 20.0°C. What is its molar mass?
Solution:
1 atm = 760 mm Hg
P = 700.0 mm Hg = 700/760 = 0.921 atm
0oC = 273.15 K
T = 20.0 oC = 20.0 + 273.15 K = 293.15 K
R = 0.082057 L⋅atm⋅K-1⋅mol-1
Ideal gas equation:
PV = nRT
Rearrange it to:
n = PV/RT = (0.921atm⋅48L)/(0.082057L⋅atm⋅K-1⋅mol-1⋅293.15 K) = 1.84 mol
M = m/n = 96.0 g : 1.84 mol = 52.2 g/ mol
Answer: 52.2 g/mol
Comments
Leave a comment