Aniline (MW93 g/mol) co-distills with water 98.2 C. The vapor pressure of water at 98.2 C is 717 mmHg. You need to isolate 10 g of aniline. What is the minimum volume of water required to isolate 10 g of aniline in a steam distillation at 760 mmHg?
With A = mass of aniline and W = mass H2O, pA = partial pressure aniline and pW = partial pressure of H2O, then
mass A/mass W = (pA*molar mass A)/pW x molar mass W)
Plug in and solve for mass W.
For pA you will want 760-43 since the pA + pW = 760 at the boiling point. mass A = 10g and solve for mass W. pW = 717 in the problem.
To calculate the minimum volume of water required to isolate 10 g of aniline in a steam distillation, we need to consider the vapor pressure of aniline at the distillation temperature.
From the given information, we know that the co-distillation of aniline with water occurs at 98.2°C, and the vapor pressure of water at this temperature is 717 mmHg. Assume that the vapor pressure of aniline at this temperature is negligible compared to the vapor pressure of water.
To calculate the minimum volume of water required, we need to use Raoult's law, which states that the total vapor pressure above a mixture of two volatile substances is equal to the sum of the partial pressures of each component. In this case, we assume aniline's vapor pressure is negligible, so the total vapor pressure is equal to the vapor pressure of water at the distillation temperature.
Let's begin by finding the mole fraction of aniline in the mixture using the given mass of aniline (10 g) and its molar mass (MW = 93 g/mol).
Mole of aniline = Mass of aniline / Molar mass of aniline
= 10 g / 93 g/mol
= 0.1075 mol
Next, we find the mole fraction of water in the mixture. Since we assume the vapor pressure of aniline is negligible, the total mole fraction will be equal to 1.
Mole of water = Total mole - Mole of aniline
= 1 - 0.1075 mol
= 0.8925 mol
Now we can use Raoult's law to calculate the partial vapor pressure of water.
Partial pressure of water = Mole fraction of water * Vapor pressure of water
= 0.8925 mol * 717 mmHg
= 638.6325 mmHg
Since we want the total pressure to be 760 mmHg, we subtract the partial pressure of water from the total pressure to find the pressure contributed by aniline.
Partial pressure of aniline = Total pressure - Partial pressure of water
= 760 mmHg - 638.6325 mmHg
= 121.3675 mmHg
Next, we'll calculate the minimum volume of water required using the ideal gas law, assuming both water and aniline are ideal gases. The equation is:
PV = nRT
Rearranging the equation, we have:
V = (nRT) / P
where V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), T is the temperature in Kelvin, and P is the pressure in atm.
Converting °C to Kelvin:
Temperature in Kelvin = 98.2°C + 273.15
= 371.35 K
Now we can calculate the minimum volume of water required:
V = (nRT) / P
= (0.8925 mol * 0.0821 L·atm/(mol·K) * 371.35 K) / (121.3675 mmHg / 760 mmHg/atm)
= 0.0865 L
Therefore, the minimum volume of water required to isolate 10 g of aniline in a steam distillation at 760 mmHg is approximately 0.0865 liters.
To determine the minimum volume of water required to isolate 10 g of aniline in a steam distillation at 760 mmHg, we need to understand the principles of steam distillation and the vapor pressure of aniline.
In steam distillation, a mixture of two immiscible liquids is heated to create vapor and then condensed to separate the desired compound from the mixture. The compound with the lower boiling point, in this case, water, vaporizes along with the desired compound, aniline, which has a higher boiling point. The vapor pressure of each component determines how much of it will evaporate at a given temperature.
To begin, we need to calculate the vapor pressure of aniline at the distillation temperature of 98.2 °C. Since this information is not provided, we can use the Clausius-Clapeyron equation to estimate it. The equation is as follows:
ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1)
Where:
P1 is the known vapor pressure (717 mmHg)
P2 is the unknown vapor pressure of aniline
ΔHvap is the enthalpy of vaporization (unknown for aniline)
R is the gas constant (0.0821 L·atm/(mol·K))
T1 is the known temperature in Kelvin (98.2 + 273.15)
T2 is the unknown temperature in Kelvin
Let's assume that the enthalpy of vaporization for aniline is similar to the enthalpy of vaporization of water (40.7 kJ/mol). Converting it to J/mol gives us:
ΔHvap = 40.7 * 1000 = 40700 J/mol
Now, let's calculate the unknown vapor pressure of aniline at 98.2 °C (371.35 K):
ln(P2/717) = -(40700/0.0821) * (1/371.35 - 1/371.5)
Solving this equation will give us the estimated vapor pressure of aniline at 98.2 °C.
Once we have the vapor pressure of aniline at 98.2 °C, we can use it to calculate the minimum volume of water required to isolate 10 g of aniline.
Since we know the pressure (760 mmHg) and the temperature (98.2 °C), we can use the Ideal Gas Law equation, assuming the ideal gas behavior for water vapor:
PV = nRT
Where:
P is the pressure (in atm)
V is the volume of water (in liters) needed
n is the number of moles of water
R is the gas constant (0.0821 L·atm/(mol·K))
T is the temperature in Kelvin (98.2 + 273.15)
We need to solve this equation for n (moles of water) to determine the minimum volume of water.
The number of moles of water (n) can be calculated using the given mass and the molar mass of water:
n = mass / molar mass
Now we have all the information needed to calculate the minimum volume of water required to isolate 10 g of aniline in a steam distillation at 760 mmHg. By following the steps outlined above, you should be able to obtain the answer.