A sample of oxygen gas at 20.0°C has a volume of 26.0L and exerts a pressure of 640.0mmHg. What is the mass of this gas?
"T = 20 + 273 = 293 \\;K \\\\\n\nV = 26.0 \\;L \\\\\n\np = 640.0 \\; mmHg = 0.8421 \\;atm"
Ideal gas law
pV = nRT
R= 0.08206 L×atm/mol×K
"n = \\frac{pV}{RT} \\\\\n\n= \\frac{0.8421 \\times 26.0}{0.08206 \\times 293} \\\\\n\n= 0.91 \\;mol \\\\\n\nm = n \\times M"
M(O2) = 32 g/mol
"m = 0.91 \\times 32 = 29.14 \\;g"
Answer: 29.14 g
Comments
Leave a comment