What is the total vapor pressure4 at 20° C of a liquid solution containing 0.30 mole fraction benzene, C6H6, and 0.70 mole fraction toluene, C6H5CH3? Assume that Raoult’s law holds for each component of the solution. The vapor pressure of pure benzene at 20° C is 75 mmHg; that of toluene at 20° C is 22 mmHg. Express your answer in mmHg.
Pbenzene = mole fraction benzene*normal vapor pressure benzene.
Ptoluene = mole fraction toluene*normal vapor pressure toluene.
Total pressure = sum of each partial pressure.
To calculate the total vapor pressure of the solution, we can use Raoult's law. According to Raoult's law, the partial vapor pressure of each component in the solution is proportional to its mole fraction.
First, we need to find the partial vapor pressures of benzene (C6H6) and toluene (C6H5CH3) in the solution.
The partial vapor pressure of benzene (Pb) can be calculated by multiplying the mole fraction of benzene (Xb) by the vapor pressure of pure benzene (Pb0).
Pb = Xb * Pb0
Substituting the given values:
Xb = 0.30 (mole fraction of benzene)
Pb0 = 75 mmHg (vapor pressure of pure benzene)
Pb = 0.30 * 75 mmHg = 22.5 mmHg
Similarly, the partial vapor pressure of toluene (Pt) can be calculated using its mole fraction (Xt) and vapor pressure of pure toluene (Pt0).
Pt = Xt * Pt0
Substituting the given values:
Xt = 0.70 (mole fraction of toluene)
Pt0 = 22 mmHg (vapor pressure of pure toluene)
Pt = 0.70 * 22 mmHg = 15.4 mmHg
Now, to find the total vapor pressure (Ptotal) of the solution, we can simply add the partial vapor pressures of benzene and toluene.
Ptotal = Pb + Pt
Ptotal = 22.5 mmHg + 15.4 mmHg = 37.9 mmHg
Therefore, the total vapor pressure of the solution at 20°C is 37.9 mmHg.
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