A mixture of ethyl iodide (b.p. 72.3) and water boils at 63.7 C. What weight of ethyl iodide would be carried over by 1 gm of steam during steam distillation
Do you have the vapor pressure at the appropriate temperature in your problem.
To determine the weight of ethyl iodide carried over by 1 gm of steam during steam distillation, we need to calculate the vapor pressure of ethyl iodide at the boiling point of the mixture.
1. Subtract the boiling point of the mixture from the boiling point of water to find the temperature difference:
Temperature difference = Boiling point of water - Boiling point of the mixture
= 100°C - 63.7°C
= 36.3°C
2. Research the vapor pressure of ethyl iodide at the temperature of the mixture's boiling point (63.7°C). For this, we can use the Antoine equation or vapor pressure tables. Let's assume the vapor pressure of ethyl iodide at 63.7°C is P.
3. Calculate the vapor pressure of water at the temperature of the mixture's boiling point (63.7°C). According to the Antoine equation for water, at 63.7°C, the vapor pressure of water is approximately 12.78 kPa.
4. Use Raoult's Law, which states that the total vapor pressure of a mixture is equal to the sum of the partial pressures of its components:
P_total = P_ethyl_iodide + P_water
Since P_water is given as 12.78 kPa and P_total is the vapor pressure of the mixture at its boiling point, we can rewrite the equation as:
P_total = P + 12.78 kPa
5. Calculate the weight of ethyl iodide carried over by 1 gm of steam using Dalton's Law of Partial Pressures, which states that the pressure exerted by each component of a mixture of gases is directly proportional to its mole fraction:
Mole fraction of ethyl iodide = P_ethyl_iodide / P_total
Weight of ethyl iodide carried over = (Mole fraction of ethyl iodide) × (1 gm of steam)
However, we need the mole fraction of ethyl iodide in terms of weight, so we can rewrite the formula as:
Weight of ethyl iodide carried over = (Mole fraction of ethyl iodide) × (Weight of steam)
The weight of steam is 1 gm.
So, to summarize:
Weight of ethyl iodide carried over by 1 gm of steam = (P_ethyl_iodide / (P + 12.78 kPa)) × 1 gm
Please note that to complete the calculation, you will need to find the vapor pressure of ethyl iodide at the boiling temperature of the mixture (63.7°C) from reference sources or experimental data.
To find the weight of ethyl iodide carried over by 1 gram of steam during steam distillation, we need to calculate the amount of ethyl iodide present in the mixture.
The boiling point of a mixture of two liquids is lower than the boiling point of either of the pure liquids. This phenomenon is known as boiling point depression. We can use Raoult's law to relate the boiling point to the composition of the mixture:
(Pure liquid vapor pressure) = (mole fraction of pure liquid)*(Vapor pressure constant)
In this case, let's assume that the mole fraction of ethyl iodide in the mixture is x, and the mole fraction of water is 1 - x. The vapor pressure constant can also be considered as the vapor pressure of pure water, since ethyl iodide is not very volatile at room temperature.
We can set up the following equation using Raoult's law:
(ethyl iodide vapor pressure at 63.7 C) = x * (vapor pressure constant)
To find the weight of ethyl iodide carried over by 1 gram of steam, we need to determine the number of moles of ethyl iodide in the vapor phase.
First, we calculate the number of moles of water in 1 gram of steam:
Number of moles of water = (mass of water) / (molar mass of water)
= 1 gram / 18.015 g/mol = 0.0555 moles
Next, we use the ideal gas law to calculate the volume of water vapor:
PV = nRT
We know that the pressure (P) is atmospheric pressure, the number of moles (n) is 0.0555, the ideal gas constant (R) is 0.0821 L·atm/(mol·K), and the temperature (T) is the boiling point of water, which can be converted to Kelvin.
Let's assume that the volume of water vapor is V liters. The equation can be rearranged as:
V = (nRT) / P
Now, we need to determine the volume of ethyl iodide vapor at the boiling point of the mixture. Since the mole fraction of ethyl iodide in the vapor phase is the same as the mole fraction of ethyl iodide in the liquid phase, we can use the ideal gas law again to calculate the volume:
V_ethyl iodide = (n_ethyl iodide * R * T) / P
Finally, we can calculate the weight of ethyl iodide carried over by 1 gram of steam:
Weight of ethyl iodide = (V_ethyl iodide / V) * (molar mass of ethyl iodide)
By substituting the values into the equations, you can find the weight of ethyl iodide carried over by 1 gram of steam during steam distillation.
You need only to know the vapor pressure of water at that b.p. which is ~ 179 mm hg, so the (vapor pressure of C2H5I = 760 - 179 )and since you know (M.W. of H2O = 18 g/mol, & M.W. C2H5I = 156 g/mol), you just apply the equation:
grams H2O M.W. H2O × P H2O
---------------------- = --------------------------------------- = 28.13 grams C2H5I
grams C2H5I M.W. C2H5I × P C2H5I