Calculating Mole Fraction of C6H6(l) in Solution

X_A)(0.526 atm) + (1-X_A)(0.188 atm)0.485 atm = 0.526X_A + 0.188 - 0.188X_A0.297 atm = 0.338X_AX_A = 0.297/0.338 = 0.879In summary, the mole fraction of C_6H_6(l) in the liquid solution is 0.879. This can be found by using the formula P_A = (X_A)(P^o _A) and substituting the given values for P_{tot}, P^o _A, and P^o _B.
  • #1
mikep
43
0
does anyone know how to do this problem?
The total vapor pressure above a solution of [tex]C_6H_6(l)[/tex] and [tex]C_7H_8(l)[/tex] is 0.485 atm. What is the mole fraction of [tex]C_6H_6(l)[/tex] in the liquid solution?
[tex]P^o C6H6(l) = 0.526 atm[/tex]
[tex]P^o C7H8(l) = 0.188 atm[/tex]

the answer is 0.879 but I'm not sure how to get that using the formula [tex]P_A = (X_A)(P^o _A)[/tex]
 
Last edited:
Chemistry news on Phys.org
  • #2
mikep said:
does anyone know how to do this problem?
The total vapor pressure above a solution of [tex]C_6H_6(l)[/tex] and [tex]C_7H_8(l)[/tex] is 0.485 atm. What is the mole fraction of [tex]C_6H_6(l)[/tex] in the liquid solution?
[tex]P^o C6H6(l) = 0.526 atm[/tex]
[tex]P^o C7H8(l) = 0.188 atm[/tex]

the answer is 0.879 but I'm not sure how to get that using the formula [tex]P_A = (X_A)(P^o _A)[/tex]

It's a direct substitution problem.

[tex]P_{tot} = P_A + P_B = (X_A)(P^o _A) + (X_B)(P^o _B) = (X_A)(P^o _A) + (1-X_A)(P^o _B)[/tex]

Now just plug in the given numbers and solve for [itex]X_A [/itex]
 
  • #3


To calculate the mole fraction of C6H6(l) in the liquid solution, first we need to calculate the total pressure of the solution, which is the sum of the vapor pressures of the individual components.

P_total = P_C6H6 + P_C7H8

Substituting the given values, we get:

P_total = 0.526 atm + 0.188 atm = 0.714 atm

Next, we can use the formula for mole fraction:

X_C6H6 = moles of C6H6 / total moles of solution

Since we are dealing with vapor pressures, we can use the ideal gas law to calculate the moles of each component:

n_C6H6 = (P_C6H6 * V) / (RT)

n_C7H8 = (P_C7H8 * V) / (RT)

Where V is the volume of the solution and R is the gas constant.

Substituting the given values, we get:

n_C6H6 = (0.526 atm * V) / (0.0821 L*atm/mol*K * 298 K) = 0.0208 mol

n_C7H8 = (0.188 atm * V) / (0.0821 L*atm/mol*K * 298 K) = 0.0075 mol

Total moles of solution = n_C6H6 + n_C7H8 = 0.0208 mol + 0.0075 mol = 0.0283 mol

Now we can calculate the mole fraction of C6H6:

X_C6H6 = 0.0208 mol / 0.0283 mol = 0.736

Therefore, the mole fraction of C6H6 in the liquid solution is 0.736.
 

1. How do you calculate the mole fraction of C6H6(l) in a solution?

The mole fraction of C6H6(l) can be calculated by dividing the moles of C6H6(l) by the total moles of all components in the solution.

2. What is the formula for calculating mole fraction?

The formula for mole fraction is Xa = na / ntotal, where Xa is the mole fraction of component a, na is the moles of component a, and ntotal is the total moles of all components.

3. How do you convert mass percent to mole fraction?

To convert mass percent to mole fraction, you will need to first calculate the moles of each component in the solution. Then, divide the moles of C6H6(l) by the total moles of all components to get the mole fraction.

4. What is the significance of calculating the mole fraction of C6H6(l) in a solution?

Calculating the mole fraction of C6H6(l) in a solution allows us to determine the amount of C6H6(l) present in relation to the total amount of all components. This is important for understanding the concentration and properties of the solution.

5. Can the mole fraction of C6H6(l) ever be greater than 1?

No, the mole fraction of any component in a solution cannot be greater than 1. This is because 1 represents 100% of the solution, and no component can make up more than 100% of the solution.

Similar threads

  • Biology and Chemistry Homework Help
Replies
5
Views
3K
  • Biology and Chemistry Homework Help
Replies
6
Views
1K
Replies
1
Views
2K
  • Biology and Chemistry Homework Help
Replies
4
Views
6K
  • Advanced Physics Homework Help
Replies
9
Views
11K
  • Biology and Chemistry Homework Help
Replies
14
Views
2K
  • Biology and Chemistry Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
995
  • Biology and Chemistry Homework Help
Replies
1
Views
7K
  • Introductory Physics Homework Help
Replies
24
Views
2K
Back
Top