26. The density of a gas at 91.45 °C and 608 torr is 1.90 g/mL. What is the molar mass of this gas?
A. 20.2
B. 37.4
C. 71.1
D. 83.8
E. 154
Pick a convenient volume of gas: 1.000 L (defined to as many significant figures as necessary to not affect any calculated answers).
Using the ideal gas law: PV = nRT or n = PV/RT
P = 608 torr×760 torr/atm = 0.800 atm
V = 1.000 L
R = 0.08206 L·atm/mole·K
T = 91.45 + 273.15 = 364.60 K
n = (0.800 atm)(1.000 L)/(0.08206 L·atm/mol·K)(364.60 K) = 0.0267 mol of the gas.
Mass of the gas = (1.000 L)×(1.90 g/L) = 1.90 g.
molar mass = mass/moles = 1.90 g/0.0267 mol = 71.1 g/mol.
Comment: this is a standard ideal gas problem.
