What is the vapor pressure of a solution composed of 3.0 moles benzene (vapor pressure 12.1 kPa) & 1 mole chloroform (vapor pressure 25.2 kPa)
vaporpressure*4parts=3parts*12.1kpa+1part*25.2kpa
divide both sides by 4 parts.
Thanks again! I have a bunch more. Your help is really appreciated.
To determine the vapor pressure of a solution composed of multiple components, you can use Raoult's Law. According to Raoult's Law, the total vapor pressure of the solution is equal to the sum of the partial pressures of each component.
Mathematically, the vapor pressure of the solution (P_total) can be calculated as:
P_total = P_1 * X_1 + P_2 * X_2 + ... + P_n * X_n
Where P_1, P_2, ..., P_n are the vapor pressures of the individual components, and X_1, X_2, ..., X_n are the mole fractions of the components in the solution.
In this case, we are given that the solution is composed of 3.0 moles of benzene and 1 mole of chloroform.
To find the mole fraction of each component, we divide the moles of each component by the total moles in the solution:
Mole fraction of benzene (X_benzene) = moles of benzene / total moles
X_benzene = 3.0 / (3.0 + 1.0) = 0.75
Mole fraction of chloroform (X_chloroform) = moles of chloroform / total moles
X_chloroform = 1.0 / (3.0 + 1.0) = 0.25
Now we can calculate the vapor pressure of the solution using Raoult's Law:
P_total = P_benzene * X_benzene + P_chloroform * X_chloroform
Substituting the given values:
P_total = 12.1 kPa * 0.75 + 25.2 kPa * 0.25
P_total = 9.075 kPa + 6.3 kPa
P_total = 15.375 kPa
Therefore, the vapor pressure of the solution composed of 3.0 moles of benzene and 1 mole of chloroform is 15.375 kPa.