Question

Question:

→ A solution has 2:3 mole ratio of hexane and benzene. If vapour pressure of hexane and benzene at 25°C are 360 mm of Hg and 300 mm of Hg respectively then the mole fraction of hexane in the vapour phase would be

1) 0.44
2) 0.55
3) 0.35
4) 0.72​

Answer 1

Answer:

pentane , = 1/(1 + 4) = 1/5 = 0.2

mole fraction of hexane , = 4/(1 + 4) = 4/5 = 0.8

vapor pressure of pure pentane at 20°C , = 400mm Hg

vapor pressure of pure hexane at 20°C , = 120mm Hg

now, total pressure of solution,

= 0.2 × 400 + 0.8 × 120

= 80 + 96

= 176 mm Hg

now mole fraction of pentane in vapor phase,

= 0.2 × 400/176 = 80/176 =0.45

Answer 2

Answer:

Mole fraction of hexane in the vapour phase ($Y_{H}$) = 0.3571

The correct option is 3.

Explanation:

TO FIND: The mole fraction of hexane in the vapour phase .

GIVEN:

       A solution has 2:3 mole ratio of hexane and benzene.

       If vapour pressure of hexane and benzene at 25°C are 360 mm of Hg and 300 mm of Hg.

                   $n_{H} : n_{B} \\2:3$

    Vapour pressure  hexane is  $P_{H}$ 360 mm

      Vapour pressure  benzene is $P_{B}$  300 mm

                    $P_{T} = P_{B}+ P_{H}$

                            $\frac{3}{5} *360+\frac{2}{5} *300\\216+120\\=336$

         

   Mole fraction of hexane in the vapour phase( $Y_{H}$)

                              $P_{H}= Y_{H} P_{T}$

                              $120=Y_{H} 336\\\frac{120}{336} =Y_{H} \\Y_{H} =0.357$

Mole fraction of hexane in the vapour phase ($Y_{H}$) = 0.3571

The project code is #SPJ2