For a solution of two liquids "A" and "B" it was proved that `P_(S)=x_(A)(p_(A)^(0)-p_(B)^(0))+p_(B)^(0)` .The resulting solution will be "

The vapour pressure of the solution of two liquids A (P^@=80mm) and B (P^@=120mm) is found to 100mm when x_(A)=0.4 .The result shows that

For an binary mixture of A and B with p_(A)^(0)ltp_(B)^(0) : x_(A)= mole fraction of A in solution x_(B)= mole fraction of B in solution y_(A)= mole fraction of A in vapour phase y_(B)= mole fraction of B in vapour phase solution

The total pressure exerted in ideal binary solution is given by P=P_(A)^(@)X_(A)+P_(B)^(@)X_(B) where P_(A)^(@)&P_(B)^(@) are the respective vapour pressure of pure components and X_(A)&X_(B) are their mole fraction in liquid phase. And composition of the vapour phase is determined with the help of Datton's law partial pressure: Y_(A)=(P_(A)^(@)X_(A))/(P) If total pressure exerted in an ideal binary solution is given by P=(5400)/(60+30Y_(A))mm of Hg. If the value of Y_(A)=0.4 then the value of X_(B) is:

The total pressure exerted in ideal binary solution is given by P=P_(A)^(@)X_(A)+P_(B)^(@)X_(B) where P_(A)^(@)&P_(B)^(@) are the respective vapour pressure of pure components and X_(A)&X_(B) are their mole fraction in liquid phase. And composition of the vapour phase is determined with the help of Datton's law partial pressure: Y_(A)=(P_(A)^(@)X_(A))/(P) If total pressure exerted in an ideal binary solution is given by P=(5400)/(60+30Y_(A))mm of Hg. The more volatile liquid is:

The total pressure exerted in ideal binary solution is given by P=P_(A)^(@)X_(A)+P_(B)^(@)X_(B) where P_(A)^(@)&P_(B)^(@) are the respective vapour pressure of pure components and X_(A)&X_(B) are their mole fraction in liquid phase. And composition of the vapour phase is determined with the help of Datton's law partial pressure: Y_(A)=(P_(A)^(@)X_(A))/(P) If total pressure exerted in an ideal binary solution is given by P=(5400)/(60+30Y_(A))mm of Hg. The value of P_(A)^(@) is:

If A and B are events such that P(A)=0.4,P(B)=0.3and P(AcupB)=0.5 then P(B'capA) equals to

If A and B are two independent events such that P(B) = (2)/(7) , P(A cup B^(c)) = 0.8 , then find P(A).

For a binary liquid solution of A and B. P_A^0 =pure vapour pressure of A . P_B^0 =pure vapour pressure of B. X_A =mole fraction of A in liquid phase. Y_A =mole fraction of A in vapour phase. {:("Column I","Column II"),((A)P_A^0gtP_B^0"[Ideal liquid solution]",(p)X_A=Y_A),((B)"Azeotropic mixture",(q)X_AltY_A),(( C)"Equimolar ideal mixture of A & B with "P_A^0ltP_B^0,(r)X_BltY_B),((D)"Equimolar ideal mixture of A & B with "P_A^0=P_B^0,(s)Y_BgtY_A),(,(t)X_B=Y_B):}

For a mixture of two volatile , completely miscible liquids A and B , with P_(A)^(@)=500 " torr and " P_(B)^(@)=800 torr , what is the composition of last droplet of liquid remaining in equilibrium with vapour ? Provided the initial ideal solution has a composition of x_(A) = 0.6 and x_(B)=0.4

If A and B are two events such that P(A) =0.6 , P(B) = 0.2 and P(A//B) = 0.5 , then P(A,B)is equal to