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Vapour pressure of C(6)H(6) and C(7)H(8)...

Vapour pressure of `C_(6)H_(6)` and `C_(7)H_(8)` mixture at `50^(circ)C` are given by `P=179X_(B)+92`, where `X_(B)` is mole fraction of `C_(6)H_(6)`. Calculate (in `mm`):
(a) Vapour pressure of pure liquids.
(b) Vapour pressure of liquid mixture obtained by mixing `936g C_(6)H_(6)` and `736g` toluene.
( c) If the vapours are removed and condensed into liquid and again brought to the temperature of `50^(circ)C`, what would be mole fraction of `C_(6)H_(6)` in vapour state ?

Text Solution

Verified by Experts

The correct Answer is:
(a) `92 mm`, (b) `199.4 mm`,
( c) `0.815, 0.185, 0.928, 0.072`;
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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?

Vapour pressure of C_(6)H_(6) and C_(7)H_(8) mixture at 50^(@)C is given by P(mm Hg) = 179 X_(B) +92 , where X_(R) is the mole fraction of C_(6)H_(6) . A solution is prepared by mixing 936g benzene and 736g toluene and if the vapours over this solution are removed and condensed into liquid and again brought to the temperature of 50^(@)C . what would be mole fraction of C_(6)H_(6) in the vapour state?

Knowledge Check

  • If the total vapour presure of the liquid mixture A and B is given by the equation: P=180X_(A)+90 then the ratio of the vapour pressure of the pure liquids A and B is given by :

    A
    `3:2`
    B
    `4:1`
    C
    `3:1`
    D
    `6:2`

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