P = 330 torr
V = 0.461 L
R = 62.364 L Torr / K mol
T = -59°C = -50 + 273.15 = 214.15 K
moles, n = PV/RT = 330*0.461/(62.364*214.15) = 0.114 mol
mass of CH₄ = moles of CH₄ x molar mass of CH₄ = (0.114 mol) x (16.04 g/mol) = 1.829 g
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Answers & Comments
P = 330 torr
V = 0.461 L
R = 62.364 L Torr / K mol
T = -59°C = -50 + 273.15 = 214.15 K
moles, n = PV/RT = 330*0.461/(62.364*214.15) = 0.114 mol
mass of CH₄ = moles of CH₄ x molar mass of CH₄ = (0.114 mol) x (16.04 g/mol) = 1.829 g