At 20°C the vapor pressure of benzene (C6H6) is 75 torr, and that of toluene (C7H8) is 22 torr. Assume that benzene and toluene form an ideal solution. (a) What is the composition... At 20°C the vapor pressure of benzene (C6H6) is 75 torr, and that of toluene (C7H8) is 22 torr. Assume that benzene and toluene form an ideal solution. (a) What is the composition in mole fractions of a solution that has a vapor pressure of 35 torr at 20°C? (b) What is the mole fraction of benzene in the vapor above the solution described in part (a)?
Understand the Problem
The question is asking for the mole fraction composition of a solution of benzene and toluene that results in a total vapor pressure of 35 torr at a given temperature. The second part asks for the mole fraction of benzene in the vapor above that same solution. To solve this, we will use Raoult's law and the ideal gas law related to vapor pressures.
Answer
(a) \( x_{C_6H_6} \approx 0.245, x_{C_7H_8} \approx 0.755 \); (b) \( y_{C_6H_6} \approx 0.529 \).
Answer for screen readers
(a) The mole fractions are ( x_{C_6H_6} \approx 0.245 ) and ( x_{C_7H_8} \approx 0.755 ).
(b) The mole fraction of benzene in the vapor is ( y_{C_6H_6} \approx 0.529 ).
Steps to Solve
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Set up Raoult's Law
Raoult's Law states that the partial vapor pressure of a component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution. For benzene ($C_6H_6$) and toluene ($C_7H_8$):
$$ P_{total} = P_{C_6H_6} + P_{C_7H_8} $$
where $$ P_{C_6H_6} = P^{\circ}{C_6H_6} \cdot x{C_6H_6} $$ $$ P_{C_7H_8} = P^{\circ}{C_7H_8} \cdot x{C_7H_8} $$
Here, ( P^{\circ}{C_6H_6} = 75 , \text{torr} ) and ( P^{\circ}{C_7H_8} = 22 , \text{torr} ).
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Expressing mole fractions
The mole fractions can be expressed as $$ x_{C_6H_6} + x_{C_7H_8} = 1 $$
Let ( x_{C_6H_6} = x ) and thus ( x_{C_7H_8} = 1 - x ).
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Substituting into the total vapor pressure equation
Substitute the expressions of the partial pressures into the total vapor pressure equation:
$$ P_{total} = (75 , \text{torr}) \cdot x + (22 , \text{torr}) \cdot (1 - x) $$
Set ( P_{total} = 35 , \text{torr} ):
$$ 35 = 75x + 22(1 - x) $$
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Solving for (x)
Simplify and solve for ( x ):
$$ 35 = 75x + 22 - 22x $$
Combine terms:
$$ 35 - 22 = 53x $$
Thus,
$$ 13 = 53x $$
Finally,
$$ x = \frac{13}{53} \approx 0.245 $$
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Finding the mole fraction of toluene
Use ( x_{C_7H_8} = 1 - x ):
$$ x_{C_7H_8} = 1 - 0.245 = 0.755 $$
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Finding the mole fraction of benzene in the vapor
To find the mole fraction of benzene in the vapor $\left( y_{C_6H_6} \right)$, use the formula:
$$ y_{C_6H_6} = \frac{P_{C_6H_6}}{P_{total}} $$
Calculate ( P_{C_6H_6} ):
$$ P_{C_6H_6} = P^{\circ}{C_6H_6} \cdot x{C_6H_6} = 75 \cdot \left(\frac{13}{53}\right) \approx 18.51 , \text{torr} $$
Thus,
$$ y_{C_6H_6} = \frac{18.51}{35} \approx 0.529 $$
(a) The mole fractions are ( x_{C_6H_6} \approx 0.245 ) and ( x_{C_7H_8} \approx 0.755 ).
(b) The mole fraction of benzene in the vapor is ( y_{C_6H_6} \approx 0.529 ).
More Information
The mole fraction reflects the proportions of each component in the solution. This type of calculation is important in understanding vapor-liquid equilibrium in mixtures. Raoult's Law is foundational in physical chemistry and is frequently used in distillation processes.
Tips
- Forgetting to total the mole fractions to 1. Always check ( x_{C_6H_6} + x_{C_7H_8} = 1 ).
- Misapplying Raoult's law; ensure you correctly identify the relationship between mole fractions and partial pressures.
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