Final answer:
The percent by mass of limonene in the steam-distillate can be calculated using Dalton's Law of Partial Pressures and the mole fractions of limonene and water, combined with their molecular weights.
Step-by-step explanation:
To calculate the percent by mass of limonene in the steam-distillate at 97.5 °C when it co-distills with water, we first need to consider the total vapor pressure of the system at this temperature, which is given as 760 mmHg. According to Dalton's Law of Partial Pressures, the total pressure is the sum of the partial pressures of limonene and water. The question provides the vapor pressure of water at this temperature as 690 mmHg. The vapor pressure of limonene can be found by subtracting the vapor pressure of water from the total pressure.
Vapor pressure of limonene = Total pressure - Vapor pressure of water
Vapor pressure of limonene = 760 mmHg - 690 mmHg = 70 mmHg
Now, we use the formula for the partial pressure of a component in a mixture: P = X × P°, where P is the partial pressure, X is the mole fraction, and P° is the standard pressure (760 mmHg). For limonene and water, we have:
Xlimonene = Plimonene / P° = 70 mmHg / 760 mmHg
Solving this gives us the mole fraction of limonene in the vapor phase. Remembering that mole fraction is the ratio of moles of limonene to the total moles in the vapor, we can calculate the mass percent of limonene in the distillate using its molecular weight:
Mass percent of limonene = (mole fraction of limonene × molecular weight of limonene) / (mole fraction of limonene × molecular weight of limonene + mole fraction of water × molecular weight of water) × 100%
Using the known molecular weight of limonene (136 g/mol) and water (18 g/mol), we can plug in the values and calculate Mass percent of limonene in the steam-distillate.