To solve the problem of determining the molecular weight and vapor pressure of the organic liquid in a total immiscible system of water and an organic liquid, we can apply Raoult's Law and some basic principles of vapor-liquid equilibrium. Let's break this down step by step.
Understanding the System
In this scenario, we have a mixture of water and an organic liquid that boils at 90°C under a pressure of 734 mmHg. We also know that the vapor pressure of water at this temperature is 526 mmHg. The distillate contains 73% of the organic liquid, which implies that the remaining 27% is water.
Applying Raoult's Law
Raoult's Law states that the partial vapor pressure of each component in a mixture is equal to the vapor pressure of the pure component multiplied by its mole fraction in the liquid phase. The total vapor pressure of the system is the sum of the partial pressures of each component:
- P_total = P_water + P_organic
Where:
- P_water = vapor pressure of water
- P_organic = vapor pressure of the organic liquid
Calculating Partial Pressures
We know:
- P_total = 734 mmHg
- P_water = 526 mmHg
Now, we can find the partial pressure of the organic liquid:
- P_organic = P_total - P_water
- P_organic = 734 mmHg - 526 mmHg = 208 mmHg
Finding Mole Fractions
Next, we need to determine the mole fractions of both components in the distillate. Given that the distillate contains 73% organic liquid and 27% water, we can express this in terms of moles:
- Let the total moles = 100 moles (for simplicity).
- Moles of organic liquid = 73 moles
- Moles of water = 27 moles
The mole fractions are then:
- X_organic = 73/100 = 0.73
- X_water = 27/100 = 0.27
Relating Vapor Pressures to Mole Fractions
According to Raoult's Law, we can express the vapor pressure of the organic liquid in terms of its mole fraction:
- P_organic = X_organic * P_organic^0
Where P_organic^0 is the vapor pressure of the pure organic liquid. We can rearrange this to find P_organic^0:
- P_organic^0 = P_organic / X_organic
- P_organic^0 = 208 mmHg / 0.73 ≈ 284.93 mmHg
Calculating Molecular Weight
To find the molecular weight of the organic liquid, we can use the formula relating vapor pressure and molecular weight:
- P_organic^0 = (R * T) / M
Where:
- R = ideal gas constant (0.0821 L·atm/(K·mol))
- T = temperature in Kelvin (90°C = 363.15 K)
- M = molecular weight of the organic liquid
Rearranging gives us:
- M = (R * T) / P_organic^0
Substituting the values:
- M = (0.0821 L·atm/(K·mol) * 363.15 K) / (284.93 mmHg * (1 atm / 760 mmHg))
- M ≈ 82.34 g/mol
Final Results
In summary, the molecular weight of the organic liquid is approximately 82.34 g/mol, and its vapor pressure at the boiling point of the mixture is about 284.93 mmHg. This analysis illustrates how vapor-liquid equilibrium principles can be applied to determine properties of components in a mixture.
