Ideal gas equation,
PV = nRT
P (250.0mL x 1 L / 1000 mL) = (0.3400 g C6H6 x 1 mol C6H6 /78.11 g C6H6)0.0821 L.atm/(mol.K) [333 K]
P = 0.476 atm
Given that the vapor pressure of benzene at this temperature is 400.0 torr
The given pressure = 0.476 atm = 0.476 atm x 760 torr / 1 atm = 361.76 torr
Pressure (361.76 torr) of benzene in the container is less than it's vapor pressure (400.0 torr). Hence all of the benzene will be in liquid state (doesnot vaprize).
As a result of a chemical reaction, 0.3400 g benzene, C_6H_6, are produced and maintained at...
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