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Q.

A solution has 1:4 molar ratio of pentane and hexane. The vapour pressure of pure hydrocarbons at 20ºC are 440 mm Hg for pentane and 120 mm of Hg for hexane. The mole fraction of pentane in vapour phase would be

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a

0.200

b

0.478

c

0.547

d

0.786

answer is B.

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Detailed Solution

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mole ratio of Pentane to hexane in liquid mixture = 1:4

$\large X_A (or) X_{pentane} =\frac{1}{1+4}=\frac{1}{5}$

$\large X_B (or) X_{hexane} =\frac{4}{1+4}=\frac{4}{5}$

X_A & X_B are molefractions in liquid phase

$\large P_{pentane}^o=440mm\;of\;Hg$

$\large P_{hexane}^o=120mm\;of\;Hg$

According to Raoult's law,

$\large P_{total}=P_A^oX_A+P_B^oX_B$

$\large P_{total}=440(\frac{1}{5})+120(\frac{4}{5})$

= (88+96)mm =184g

In vapour phase ,

According to Dalton's law

$\large P_A=P_{total}\times Y_A\;\;\;;\;\;\;P_B=P_{total}\times Y_B$

Y_A & Y_B are the mole fractions of A & B in vapour phase

$\large P_{pentane}=P_{total}\times Y_{pentane}$

$\large Y_{pentane}=\frac{P_{pentane}}{P_{total} }=\frac{440(\frac{1}{5})}{184}=\frac{88}{184}=0.478$

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