$(a)\;0.23\qquad(b)\;0.19\qquad(c)\;0.39\qquad(d)\;0.30$

$P_T = P_{O_2} + P_{CP}$

$\;\;\;\;\;=570+170$

$\;\;\;\;\;\;=740 \;torr$

Partial pressure of cyclopentane = $P_T\times$ mole fraction of cyclopentane

Partial pressure of $O_2$ = $P_T\times$ mole fraction of $O_2$

$\Large\frac{P_{CP}}{P_{O_2}} = \frac{P_T \times \text{mole fraction of cyclopentane}}{P_T \times \text{mole fraction of } O_2} = \frac{moles\; of\; cyclopentane}{moles\;of\;O_2}$

$\therefore \large\frac{Mole\;of\;cyclopentane}{Mole\; of\;O_2} = \large\frac{P_{CP}}{P_{O_2}}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\large\frac{170}{570}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=0.30$

Hence answer is (d)

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