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# How do you calculate the χacetone and χcyclohexane in the vapor above the solution. P°acetone = 229.5 torr and P°cyclohexane = 97.6 torr?

## A solution is prepared by mixing 4.40 mol of acetone with 1.45 mol of cyclohexane at 30°C.

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#### Explanation

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#### Explanation:

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Mar 21, 2018

Well by definition....the vapour pressure of EACH component is PROPORTIONAL to ${\chi}_{\text{component} \left(g\right)}$...

${P}_{\text{component" = chi_("component"(g))P_"solution}}$

#### Explanation:

Well in solution, the $\text{mole fraction}$ is

chi_("component"(l))="moles of component"/"Total moles in solution"

So,

chi_("acetone"(l))="moles of acetone"/"moles of acetone + moles of cyclohexane"...

$= \frac{4.40 \cdot m o l}{4.40 \cdot m o l + 1.45 \cdot m o l} = 0.752 .$

And clearly....${\chi}_{\text{cyclohexane"(l))=1-chi_("acetone} \left(l\right)} = 1 - 0.752 = 0.248$.

And so

P_"acetone"=chi_("acetone"(l))xx229.5*mm*Hg=0.752xx229.5*mm*Hg=172.6*mm*Hg

P_"cyclohexane"=chi_("cyclohexane"(l))xx97.6*mm*Hg=0.248xx97.6*mm*Hg=24.2*mm*Hg

${P}_{\text{solution"=P_"acetone"+P_"cyclohexane}} = \left\{172.6 + 24.2\right\} \cdot m m \cdot H g$

$\equiv 196.8 \cdot m m \cdot H g$

But in the vapour phase, the vapour pressure of EACH component is PROPORTIONAL to ${\chi}_{\text{component} \left(g\right)}$...

${P}_{\text{component" = chi_("component"(g))P_"solution}}$

As always, the vapour pressure is ENRICHED with respect to the more volatile component....here acetone...

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