| Direction: Two liquids A and B have the same molecular weights and for man ideal solution. The solution a/composition \[{{X}_{A}}\] has the vapour pressure 700 mm Hg at \[80{}^\circ C\]. The above solution is distilled without reflux till 3/4 of the solution is collected as condensate. The composition of the condensate is\[X{{'}_{A}}=0.75\] and that of residue is \[{{X}_{A}}=0.3\]. The vapour pressure of the residue at \[80{}^\circ C\] is 600 mm. |
A) 807.41 mm of Hg
B) 511.11 mm of Hg
C) 707.41 mm of Hg
D) 207.41 mm of Hg
Correct Answer: B
Solution :
Let total moles of liquid taken initially \[=x\] Condensate \[=\frac{3}{4}\,x\] and Residue \[=\frac{1}{4}\,x\] Moles of A in condensate \[=0.75\times \frac{3}{4}x=\frac{9}{16}x\] Moles of A in residue \[=0.3\times \frac{1}{4}=\frac{3}{40}x\] Total moles of \[A=\,\left( \frac{9}{16}+\frac{3}{40} \right)x\] Mole fraction of A in original sample \[=\frac{51}{80}=0.6375\] (Neglecting moles of vapour as very small) Total pressure over original solution= 700 mm Hg \[700=p_{A}^{o}\,\left( \frac{51}{80} \right)+p_{B}^{o}\,\left( \frac{29}{80} \right)\] ?(i) Total pressure above residue = 600 mm Hg \[600=p_{A}^{o}\,\left( \frac{3}{10} \right)\,+p_{B}^{o}\,\left( \frac{7}{10} \right)\] ?(ii) From (i), \[56000=51\,p_{A}^{o}+29\,p_{B}^{o}\] From (ii), \[[6000=3{{p}^{o}}+7p_{B}^{o}\times 17\] \[102000=51\,p_{A}^{o}+119\,p_{B}^{o}\] \[\therefore \] \[46000=90\,p_{B}^{o}\] \[p_{B}^{o}=511.11\,\,mm\,\,of\,\,Hg\] Putting value of \[p_{B}^{o\,}\,in\,(i)\] \[700=p_{A}^{o}\,\left( \frac{51}{80} \right)\,+\frac{4600}{9}\times \frac{29}{80}\] \[p_{A}^{o}\,=\,\left[ 700-\frac{4600}{9}\times \frac{29}{80} \right]\times \frac{80}{51}\] \[=807.41\,\,mm\,\,of\,\,Hg\]
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