A vapor pressure of toluene of 21 mm Hg, at 760 mm Hg, what is the equilibrium concentration for toluene?
Do you have a constant for toluene? (Henry's Law constant?)
p = Kc*C
p/Kc = C
no
If you have an answer to the problem I can calculate the constant.
To find the equilibrium concentration of toluene, we need to first understand what vapor pressure represents. Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid phase at a given temperature.
In this case, the vapor pressure of toluene is given as 21 mm Hg, at a total pressure of 760 mm Hg. This means that at this temperature, toluene is in equilibrium with its vapor phase, and the partial pressure of toluene in the system is 21 mm Hg.
To find the equilibrium concentration of toluene, we can use the ideal gas law, which states that the pressure of a gas (in this case, the vapor pressure of toluene) is directly proportional to the concentration of the gas (in this case, the equilibrium concentration of toluene) and the temperature.
The ideal gas law equation is given as: PV = nRT
Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of the gas
R is the ideal gas constant
T is the temperature in Kelvin
Since we are dealing with the vapor phase of toluene, we can assume that the volume is constant and cancel it out from the equation. We can also substitute the given values for P and solve for n (the number of moles of toluene).
Here's the modified equation: n = PV / RT
Given:
P (pressure) = 21 mm Hg
V (volume) = constant (for simplicity, we can assume it to be 1 liter)
R (ideal gas constant) = 0.0821 L atm/(mol K) (which can also be expressed as 62.36 L mmHg/(mol K))
T (temperature) = not given
Since the temperature is not given, we cannot calculate the equilibrium concentration of toluene without this information. The equilibrium concentration is directly dependent on the temperature.