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 (N−5.5) is________

Question

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 (N−5.5) is________

Infinity Learn · Accepted Answer

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

$x A \to A_{x} (S) t = \infty 8 - 8 = 0 \frac{8}{x} m o l \Rightarrow 300 = x_{B} .420 = \frac{5}{5 + \frac{8}{x}} \times 420$

$\Rightarrow x = 4$ , i.e Associates into tetramer $A_{4}$ . Now, at 1.0h.

$\Rightarrow p = 360 = X_{A} p^{o} (A) + X_{B} p^{o} (B) = (\frac{8 - α}{13 - 0.75 α}) 340 + (\frac{5}{13 - 0.75 α}) 420$

Solving for $α = 2$
Total moles at 1.0hr $= 13 - 0.75 α = 11.5$

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