How many moles of argon are there in 20.0 L, at 25C and 98.8kPa?



Answer:
98.8 kPa = 98800 Pa

101325 Pa = 1atm

98800 / 101325 = 0.975 atm

T = 25 + 273 = 298 K

pV = nRT

n = pV / RT = 0.975 x 20 / 0.0821 x 298 = 0.797 moles
2.36

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