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Answer :
To solve the problem, we use Raoult's law and the definition of mole fraction in the vapor phase.
1. First, determine the total number of moles in the solution:
[tex]$$
n_{\text{total}} = n_{\text{cyclohexane}} + n_{\text{acetone}} = 1.90 + 2.60 = 4.50 \text{ moles}.
$$[/tex]
2. Next, calculate the mole fractions of cyclohexane and acetone in the liquid solution:
[tex]$$
x_{\text{cyclohexane}} = \frac{1.90}{4.50} \approx 0.422, \quad x_{\text{acetone}} = \frac{2.60}{4.50} \approx 0.578.
$$[/tex]
3. According to Raoult's law, the partial pressure of each component is given by:
[tex]$$
P_i = x_i \, P_i^\circ,
$$[/tex]
where [tex]$P_i^\circ$[/tex] is the vapor pressure of the pure component. Therefore, the partial pressures are:
[tex]$$
P_{\text{cyclohexane}} = 0.422 \times 97.6 \, \text{torr} \approx 41.21 \, \text{torr},
$$[/tex]
[tex]$$
P_{\text{acetone}} = 0.578 \times 229 \, \text{torr} \approx 132.31 \, \text{torr}.
$$[/tex]
4. The total vapor pressure of the system is the sum of the partial pressures:
[tex]$$
P_{\text{total}} = P_{\text{cyclohexane}} + P_{\text{acetone}} \approx 41.21 + 132.31 = 173.52 \, \text{torr}.
$$[/tex]
5. Finally, the mole fraction of cyclohexane in the vapor phase is given by:
[tex]$$
y_{\text{cyclohexane}} = \frac{P_{\text{cyclohexane}}}{P_{\text{total}}} \approx \frac{41.21}{173.52} \approx 0.237.
$$[/tex]
Thus, the mole fraction of cyclohexane in the vapor is approximately [tex]$\boxed{0.237}$[/tex].
1. First, determine the total number of moles in the solution:
[tex]$$
n_{\text{total}} = n_{\text{cyclohexane}} + n_{\text{acetone}} = 1.90 + 2.60 = 4.50 \text{ moles}.
$$[/tex]
2. Next, calculate the mole fractions of cyclohexane and acetone in the liquid solution:
[tex]$$
x_{\text{cyclohexane}} = \frac{1.90}{4.50} \approx 0.422, \quad x_{\text{acetone}} = \frac{2.60}{4.50} \approx 0.578.
$$[/tex]
3. According to Raoult's law, the partial pressure of each component is given by:
[tex]$$
P_i = x_i \, P_i^\circ,
$$[/tex]
where [tex]$P_i^\circ$[/tex] is the vapor pressure of the pure component. Therefore, the partial pressures are:
[tex]$$
P_{\text{cyclohexane}} = 0.422 \times 97.6 \, \text{torr} \approx 41.21 \, \text{torr},
$$[/tex]
[tex]$$
P_{\text{acetone}} = 0.578 \times 229 \, \text{torr} \approx 132.31 \, \text{torr}.
$$[/tex]
4. The total vapor pressure of the system is the sum of the partial pressures:
[tex]$$
P_{\text{total}} = P_{\text{cyclohexane}} + P_{\text{acetone}} \approx 41.21 + 132.31 = 173.52 \, \text{torr}.
$$[/tex]
5. Finally, the mole fraction of cyclohexane in the vapor phase is given by:
[tex]$$
y_{\text{cyclohexane}} = \frac{P_{\text{cyclohexane}}}{P_{\text{total}}} \approx \frac{41.21}{173.52} \approx 0.237.
$$[/tex]
Thus, the mole fraction of cyclohexane in the vapor is approximately [tex]$\boxed{0.237}$[/tex].
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