Answer to Question #349339 in General Chemistry for Happy

Question

96.0 g. of a gas occupies 48.0 L at 700.0 mm Hg and 20.0°C. What is its molar mass?

Answer 1

Solution:

1 atm = 760 mm Hg

P = 700.0 mm Hg = 700/760 = 0.921 atm

0^oC = 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