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InterviewSolution
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Vapour pressure of C_6H_6 and C_7H_8 mixture at 50^@C are given by: P=179X_B+92," where "X_B" is mole fraction of "C_6H_6. Calculate (in mm): (A) Vapour pressure of pure liquids. (B) Vapour pressure of liquid mixture obtained by mixing 936 g C_6H_6 anf 736 g toluene. (C) If the vapours are removed and condensed into liquid and again brought to the temperature of 50^@C, what would be mole fraction of C_6H_6 in vapour state? |
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Answer» Solution :(A) Given, `P=179X_B+92` For pure, `C_6H_6,X_B=1` `:.P_B^@=179+92=271mm` For pure,`C_7H_8,X_B=0` `:.P_T^@=179xx0+92=92mm` (B) Now`PM=P_B^@X_B+P_T^@X_T` `271xx12/(12+8)+92xx8/(12+8)=162.6+36.8` = 199.4 mm. MOLES of `C_6H_6=936/78=12` Moles of `C_7H_8=736/92=8` (C) Now mole fraction of `C_6H_6` in VAPOUR phase of intial mixture`(X_T^/)` `(X_T^/)=P_B^(/)/P_M=162.6/199.4=0.815` Moles fraction of `C_7H_8` in vapour phase of intial mixture `(X_T^/)` `(X_T^/)=P_T^(/)/P_M=36.8/199.4=0.185` These vapours are taken out and condensed into liquid. The liquid is again BROUGHT to `50^@` to get again vapour-liquid equilibrium. Thus, mole fraction of `C_6H_6` in vapour phase of initial mixture = Mole fraction of `C_6H_6` in liquid phase on II mixture `X_B^(/)` Similarly, mole fraction of `C_7H_8` in vapour phase of initial mixture Mole fraction of `C_7H_8` in liquid phase on II mixture `X_T^(/)` New `P_M=P_B^(/)+P_T^(/)` THEREFORE, new `P_M=P_B^@X_B^(/)+P_T^@X_T^(/)` `271xx0.815+92xx0.185` mm `220.865+17.02=237.885` mm `:.` New mole fraction of `C_6H_6` in vapour phase `("New"P_B^(/))/("New"P_M)=220.865/237.885=0.928` `:.` New mole fraction of `C_7H_8` in vapour phase = 0.072 |
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