The vapour pressure of two pure liquids A and B that form an ideal solution, are 400 and 800 mm of Hg respectively at a temperature t° C. The mole fraction of A in a solution of A and B whose boiling point is t° C will be

Question

The vapour pressure of two pure liquids A and B that form an ideal solution, are 400 and 800 mm of Hg respectively at a temperature t° C. The mole fraction of A in a solution of A and B whose boiling point is t° C will be (A) 0.4 (B) 0.8 (C) 0.1 (D) 0.2

Tardigrade Admin · Accepted Answer

V.P. of solution at t° C=760 mm [at b.p., V.P. of solution = atmospheric pressure] Thus =PA ° ⋅ xA+PB ° xB or P=PA ° ⋅ xA+PB ° ⋅(1-xA) [∵ xA+xB=1] or 760=400 XA+800(1-XA) [∵ P=760 mm of Hg ] or -800+760=-400 xA or -40=-400 xA or xA=(40/400)=0.1 Thus mole fraction in solution is 0.1

Q. The vapour pressure of two pure liquids A and B that form an ideal solution, are 400 and 800mm of Hg respectively at a temperature t∘C. The mole fraction of A in a solution of A and B whose boiling point is t∘C will be

0.4

0.8

0.1

0.2

Q. The vapour pressure of two pure liquids $A$ and $B$ that form an ideal solution, are $400$ and $800 mm$ of $H g$ respectively at a temperature $t^{\circ} C$ . The mole fraction of $A$ in a solution of $A$ and $B$ whose boiling point is $t^{\circ} C$ will be