What volume is occupied by 0.25 grams of oxygen gas measured at 25.0 deg C
and 100.66 kPa? ( 5 marks)
( universal gas constant= 8.314 Kpa. L mol-1 K-1)
p = 100.66 kPa
T = 25 + 273 = 298 K
M(O2) = 32 g/mol
m = 0.25 g
"n = \\frac{m}{M} \\\\\n\nn(O_2) = \\frac{0.25}{32} = 0.0078 \\;mol"
Ideal Gas Law
"pV = nRT \\\\\n\nV = \\frac{nRT}{p} \\\\\n\n= \\frac{0.0078 \\times 8.314 \\times 298}{100.66} \\\\\n\n= 0.192 \\;L"
Answer: 192 mL
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