Calculating Mole Fraction of C6H6(l) in Solution

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SUMMARY

The mole fraction of C6H6(l) in a solution with a total vapor pressure of 0.485 atm can be calculated using Raoult's Law. Given the vapor pressures of pure components, P^o C6H6(l) = 0.526 atm and P^o C7H8(l) = 0.188 atm, the formula P_A = (X_A)(P^o_A) is applied. By substituting the known values into the equation P_tot = (X_A)(P^o_C6H6) + (1-X_A)(P^o_C7H8), the mole fraction X_A is determined to be 0.879.

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  • Understanding of Raoult's Law
  • Basic knowledge of vapor pressure concepts
  • Familiarity with mole fraction calculations
  • Ability to perform algebraic substitutions
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  • Learn about vapor pressure and its significance in solutions
  • Explore mole fraction calculations in different contexts
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Chemistry students, chemical engineers, and anyone involved in solution chemistry or vapor-liquid equilibrium studies will benefit from this discussion.

mikep
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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]
 
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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]
 

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