A mixture of acetone and ethyl acetate of unknown ratio begins to boil at 65 degrees celcius (760 torr). What is the composition of this binary mixture in terms of the mole fraction of acetone and that of ethyl acetate?
To find the composition of the binary mixture in terms of mole fraction, we need to use the concept of Raoult's Law for liquid mixtures. Raoult's Law states that the vapor pressure of a component in a mixture is equal to the mole fraction of that component times its vapor pressure in the pure state.
Let x1 be the mole fraction of acetone (component 1) and x2 be the mole fraction of ethyl acetate (component 2). Given that the boiling point of the mixture is 65°C at 760 torr (1 atm), we can find the vapor pressures of the pure components at that temperature.
The normal boiling point of acetone is 56.05°C (329.20 K) and its vapor pressure at this temperature is 1 atm (760 torr). The normal boiling point of ethyl acetate is 77.1°C (350.25 K) and its vapor pressure at this temperature is also 1 atm (760 torr).
We can use the Antoine equation to find the vapor pressure of each component at 65°C:
log(P) = A - (B / (C + T))
For acetone:
A = 7.02447, B = 1161.0, C = 224
For ethyl acetate:
A = 7.03119, B = 1225.532, C = 202.487
Plugging the values at 65°C (338.15 K):
log(P1) = 7.02447 - (1161 / (224 + 338.15))
P1 = 242.7 torr
log(P2) = 7.03119 - (1225.532 / (202.487 + 338.15))
P2 = 76.4 torr
According to Raoult's Law:
Px = xi * Pxi
For acetone: P1 = x1 * P1
For ethyl acetate: P2 = x2 * P2
Since the total pressure is 760 torr:
P1 + P2 = 760
x1 * P1 + x2 * P2 = 760
We also know that the mole fractions should add up to 1:
x1 + x2 = 1
Solving the system of equations:
x1 * 242.7 + x2 * 76.4 = 760
x1 + x2 = 1
x2 = 1 - x1
Plugging the second equation into the first:
x1 * 242.7 + (1 - x1) * 76.4 = 760
Solving for x1:
x1 = (760 - 76.4) / (242.7 - 76.4)
x1 ≈ 0.8513
Now using the second equation:
x2 = 1 - 0.8513 = 0.1487
Therefore, the composition of the binary mixture is approximately 85.13% mole fraction of acetone and 14.87% mole fraction of ethyl acetate.
To determine the composition of the binary mixture of acetone and ethyl acetate, we can use Raoult's law, which states that the vapor pressure of a component in an ideal solution is directly proportional to its mole fraction in that solution.
Given:
Boiling point of the mixture = 65 degrees Celsius
Total pressure = 760 torr
To find the composition of the mixture, we need to make the assumption that the two components obey Raoult's law and that the boiling point is purely determined by the composition.
Step 1: Convert the boiling point to Kelvin.
65 degrees Celsius + 273.15 = 338.15 K
Step 2: Look up the vapor pressure of pure components at their boiling points.
The vapor pressure of acetone at its boiling point is 1 atm (760 torr).
The vapor pressure of ethyl acetate at its boiling point is 0.57 atm (433.2 torr).
Step 3: Calculate the mole fraction of acetone.
According to Raoult's law, the mole fraction of acetone in the solution is equal to the ratio of the partial pressure of acetone to the total pressure.
The partial pressure of acetone can be calculated using Dalton's law of partial pressures:
Partial pressure of acetone = Vapor pressure of acetone * mole fraction of acetone.
Since we don't know the mole fraction of acetone, let's assume it to be x.
Partial pressure of acetone = x * 1 atm
Step 4: Calculate the mole fraction of ethyl acetate.
Using the same logic as in Step 3, we can calculate the partial pressure of ethyl acetate and, assuming it to be y, we get:
Partial pressure of ethyl acetate = y * 0.57 atm
Step 5: Applying the law of partial pressures.
Using the information given, we can write the following equation:
Partial pressure of acetone + Partial pressure of ethyl acetate = Total pressure
x * 1 atm + y * 0.57 atm = 1 atm
Step 6: Solving the equation.
We know that the total pressure is 760 torr, which is equivalent to 1 atm. We can use this equivalency to solve for either x or y. Let's solve for y:
y = (1 - x) * (0.57/1)
Step 7: Find the value of x.
Substituting the value of y from Step 6 into the equation from Step 5, we have:
x * 1 atm + (1 - x) * (0.57/1) * 1 atm = 1 atm
Simplifying the equation, we get:
x + 0.57 - 0.57x = 1
0.43x = 0.43
x = 1
Step 7: Calculate the mole fraction of ethyl acetate (y) using the value of x.
From Step 6, we found that y = (1 - x) * (0.57/1)
Substituting the value of x, we get:
y = (1 - 1) * (0.57/1)
y = 0
Therefore, the composition of the binary mixture in terms of the mole fraction of acetone (x) and ethyl acetate (y) is x = 1 and y = 0.
To determine the composition of the binary mixture, we can use Raoult's Law, which states that the partial 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.
Let's assume that the vapor pressures of acetone and ethyl acetate at 65 degrees Celsius (760 torr) are P_acetone and P_ethylacetate, respectively.
According to Raoult's Law, we can write two equations:
P_acetone = X_acetone * P^0_acetone
P_ethylacetate = X_ethylacetate * P^0_ethylacetate
Where X_acetone and X_ethylacetate are the mole fractions of acetone and ethyl acetate, respectively, and P^0_acetone and P^0_ethylacetate are their respective vapor pressures at 65 degrees Celsius.
We know that the total pressure of the mixture is 760 torr. Thus, we can write the following equation:
P_acetone + P_ethylacetate = 760 torr
Substituting the expressions from the first two equations into the third equation, we get:
X_acetone * P^0_acetone + X_ethylacetate * P^0_ethylacetate = 760 torr
Since we don't have the values of P^0_acetone and P^0_ethylacetate, we cannot solve for X_acetone and X_ethylacetate directly.
However, we can make an assumption that the vapor pressure of the mixture is the weighted average of the vapor pressures of acetone and ethyl acetate. Mathematically, this can be expressed as:
P_mixture = X_acetone * P^0_acetone + X_ethylacetate * P^0_ethylacetate
Since the mixture begins to boil at 65 degrees Celsius, the vapor pressure of the mixture at this temperature is equal to the external pressure (760 torr). Therefore, we can rewrite the equation as:
760 torr = X_acetone * P^0_acetone + X_ethylacetate * P^0_ethylacetate
Now, we need values of the vapor pressures of acetone and ethyl acetate at 65 degrees Celsius to plug into the equation. These values are typically provided in vapor pressure charts or found in databases.
Once you have the values for P^0_acetone and P^0_ethylacetate at 65 degrees Celsius (760 torr), you can solve the equation for the mole fractions X_acetone and X_ethylacetate.
Remember that mole fractions are dimensionless values ranging from 0 to 1, and their sum should be equal to 1 in a binary mixture.