P = 0.881721 atm
T= 125.0 oC
= (125.0+273) K
= 398 K
Molar mass of CCl4,
MM = 1*MM(C) + 4*MM(Cl)
= 1*12.01 + 4*35.45
= 153.81 g/mol
Lets derive the equation to be used
use:
p*V=n*R*T
p*V=(mass/molar mass)*R*T
p*molar mass=(mass/V)*R*T
p*molar mass=density*R*T
Put Values:
0.881721 atm *153.81 g/mol = density * 0.08206 atm.L/mol.K *398.0
K
density = 4.1524 g/L
Answer: 4.15 g/L
what is the density of (CCl4) vapor at 0.881721 atm and 125 celcius? (in g/L)
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