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Q.

A and B are two volatile liquids with their vapour pressure:  PA0=340mm  and  PB0=420mm  of  Hg  at  25oC.To flask containing 8.0 mole of A, 5.0 moles of B was added. As soon as B was added, A starts associating to form a non-volatile substance Ax(solid)  which is soluble in both liquids A and B. The vapour pressures of solutions measured at the end of one hour and after a very long time were found to be 360mm and 300mm of Hg respectively. The total number of moles of A, B and  Ax(solid) at the end one hour is  N then (N5.5) is________

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answer is 6.

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Detailed Solution

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After a very long time, A will be completely associated as. 

xAAx(S) t=           88=0         8xmol 300=xB.420=55+8x×420

x=4 , i.e Associates into tetramer A4. Now, at 1.0h.

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p=360=XApo(A)+XBpo(B) =(8α130.75α)340+(5130.75α)420

Solving for  α=2
Total moles at 1.0hr  =130.75α=11.5

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