Answer:
[tex] \chi_{c(g)} = 0.235 [/tex]
Explanation:
The mole fraction of cyclohexane in the vapor [tex] \chi_{c(g)}[/tex] is:
[tex] \chi_{c(g)} = \frac{P_{c}}{P_{T}} [/tex] (1)
where [tex]P_{c}[/tex]: is the partial pressure of cyclohexane and [tex] P_{T}[/tex]: is the total pressure.
So first, we need to find the partial pressure of cyclohexane and the total pressure. To do that, we can use Raoult's Law:
[tex] P_{T} = P_{c} + P_{a} = \chi_{c}*P_{c}^{\circ} + \chi_{a}*P_{a}^{\circ} [/tex] (2)
where Pc and Pa: are the partial pressures of cyclohexane and acetone, respectively, χc and χa: are the mole fractions of cyclohexane and acetone, respectively, and Pc⁰ = 97.6 torr and Pa⁰ = 229.5 torr.
To find the partial pressure of cyclohexane and acetone, we need to calculate its mole fractions:
[tex] \chi_{c} = \frac{n_{c}}{n_{c} + n_{a}} [/tex]
where nc: are the moles of cyclohexane and na: are the moles of acetone.
[tex] \chi_{c} = \frac{1.90 mol}{1.90 mol + 2.60 mol} = 0.42 [/tex]
[tex] \chi_{a} = \frac{n_{a}}{n_{c} + n_{a}} = \frac{2.60 mol}{1.90 mol + 2.60 mol} = 0.58 [/tex]
Now, the total pressure can be calculated using equation (2):
[tex] P_{T} = \chi_{c}*P_{c}^{\circ} + \chi_{a}*P_{a}^{\circ} = 0.42*97.6 torr + 0.58*229.5 torr = 40.99 torr + 133.11 torr= 174.10 torr [/tex]
Finally, the mole fraction of cyclohexane in the vapor (equation 1) is:
[tex] \chi_{c(g)} = \frac{P_{c}}{P_{T}} = \frac{40.99 torr}{174.10 torr} = 0.235 [/tex]
I hope it helps you!
