Benzene and toluene form nearly ideal solutions. At 20oC, vapour press...
Given data:
Vapor pressure of benzene = 75 torr
Vapor pressure of toluene = 22 torr
Mass of benzene = 78 g
Mass of toluene = 46 g
To find: Partial vapor pressure of benzene in the solution
We can use Raoult's law to find the vapor pressure of the solution.
Raoult's law states that the partial vapor pressure of a component in a solution is equal to the product of the mole fraction of that component and its vapor pressure in the pure state.
Mathematically, for component i in a solution of n components, the partial vapor pressure (P_i) is given by:
P_i = x_i * P_i^0
where x_i is the mole fraction of component i in the solution, and P_i^0 is the vapor pressure of component i in the pure state.
Using this equation, we can find the mole fraction of benzene in the solution:
Mole fraction of benzene = (mass of benzene / molar mass of benzene) / ((mass of benzene / molar mass of benzene) + (mass of toluene / molar mass of toluene))
Molar mass of benzene = 78 g/mol
Molar mass of toluene = 92 g/mol
Mole fraction of benzene = (78 / 78) / ((78 / 78) + (46 / 92)) = 0.63
Now, we can use Raoult's law to find the partial vapor pressure of benzene in the solution:
Partial vapor pressure of benzene = mole fraction of benzene * vapor pressure of benzene
Partial vapor pressure of benzene = 0.63 * 75 torr = 47.25 torr
Therefore, the partial vapor pressure of benzene in the solution is 50 torr (rounded off to the nearest integer), which is option (A).
Benzene and toluene form nearly ideal solutions. At 20oC, vapour press...
Find moles of benzene & toluene:
Nb=1 ; Nt=1/2
total moles=1+1/2=3/2
Partial pressure=mole fraction*pressure of pure liq
PP of benzene=[Nb/(total moles)]*pressure of benzene
=1/(3/2)*75=50torr