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Oxygen and cyclopentane at partial pressure of 570 torr and 170 torr respectively are mixed in a gas cylinder . The ratio of the number of moles of cyclopentane to the number of moles of oxygen is:


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$P_T = P_{O_2} + P_{CP}$
$\;\;\;\;\;\;=740 \;torr$
Partial pressure of cyclopentane = $P_T\times$ mole fraction of cyclopentane
Partial pressure of $O_2$ = $P_T\times$ mole fraction of $O_2$
$\Large\frac{P_{CP}}{P_{O_2}} = \frac{P_T \times \text{mole fraction of cyclopentane}}{P_T \times \text{mole fraction of } O_2} = \frac{moles\; of\; cyclopentane}{moles\;of\;O_2}$
$\therefore \large\frac{Mole\;of\;cyclopentane}{Mole\; of\;O_2} = \large\frac{P_{CP}}{P_{O_2}}$
Hence answer is (d)
answered Mar 6, 2014 by sharmaaparna1
edited Mar 18, 2014 by mosymeow_1

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