Limone (MW 136g/mol) is a pleasant smelling liquid found in lemon and orange peels. The boiling point of limonene is 175 ˚C, but it co-distills with water at 97.5 ˚C. If the vapour pressure of water at 97.5˚C is 690 mmHg, what percent (by mass) of the steam-distillate is limonene? (Assuming that the distillation is occurring at 760 mmHg)

Ptotal= Pa+ Pb
760 = 690 + Pb
Pb = 70
Don't know what to do from here

To find the percent by mass of limonene in the steam-distillate, we need to understand the concept of vapor pressure and its relationship to the composition of a mixture. Vapor pressure is the pressure exerted by the gaseous phase of a substance in equilibrium with its liquid or solid phase at a given temperature.

In this case, we are given that the distillation is occurring at a total pressure of 760 mmHg and the vapor pressure of water is 690 mmHg at the boiling point of 97.5 ˚C. The difference in these pressures represents the vapor pressure of limonene.

To find the vapor pressure of limonene, we subtract the vapor pressure of water from the total pressure:
Pb = Ptotal - Pa
Pb = 760 mmHg - 690 mmHg
Pb = 70 mmHg

Now that we have the vapor pressure of limonene, we can use Raoult's Law to calculate the mole fraction of limonene in the steam-distillate. Raoult's Law states that the vapor pressure of a component in a mixture is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture.

Using Raoult's Law, we have:
Pb = Xb * P°b
where
Pb is the vapor pressure of limonene (70 mmHg),
Xb is the mole fraction of limonene, and
P°b is the vapor pressure of limonene at its boiling point.

We can rearrange the equation to solve for Xb:
Xb = Pb / P°b

To find P°b, the vapor pressure of limonene at its boiling point, we can assume that limonene behaves ideally and use the Clausius-Clapeyron equation:
ln(P°b / P°a) = (ΔHvap / R) * (1 / T°a - 1 / T°b)

where
P°a is the vapor pressure of limonene at a known temperature, which we don't have,
ΔHvap is the heat of vaporization of limonene (also not provided),
R is the ideal gas constant, and
T°a and T°b are the known temperatures corresponding to P°a and P°b, respectively.

Since we don't have the necessary data to directly calculate P°b or ΔHvap, we'll have to make a simplified assumption. Let's assume that limonene behaves ideally and that the heat of vaporization is constant over the temperature range. We can then use the boiling point of limonene (175 ˚C) to estimate its vapor pressure at that temperature.

Using this assumption, we can calculate Xb and then convert it to percent by mass:
Xb = 70 mmHg / P°b
percent mass of limonene = Xb * molar mass of limonene / (Xb * molar mass of limonene + (1 - Xb) * molar mass of water) * 100

It's important to note that these calculations are based on assumptions and estimations, and the exact result may vary depending on the accuracy of the assumptions and values used.