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Benzene and toluene form a nearly ideal ...

Benzene and toluene form a nearly ideal solution. At a certain temperature, calculate the vapour pressure of solution containing equal moles of two substances `("Given ",p_("toluene")^0=55mm" "Hg,p_("benzene")^0=150mm" "Hg]`

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As the number of moles of toluene and benzene are equal, `x_("toluene"),x_("benzene")=0.5`
`:.` According to Raoult's law, the vapour pressure of the solution, `P=x_("toluene")*P_("toluene")^0+x_("benzene")*P_("benzene")^0`
`=(0.5xx55+0.5xx150)mm" "Hg`
`=(27.5+75)mm" "Hg=102.5mm" "Hg`
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Knowledge Check

  • At 40^@C , the vapour pressure of pure liquids, benzene and toluene, are 160 mm Hg and 60 mm Hg respectively. At the same temperature, the vapour pressure of an equimolar solution of the two liquids, assuming the ideal solution should be

    A
    140 mmHg
    B
    110 mm Hg
    C
    220 mm Hg
    D
    100 mm Hg

  • Two liquids A and B form an ideal solution. What is the vapour pressure of solution containing 2 moles of A and 3 moles of B at 300 K? [Given : At 300 K, vapour pressure of pure liquid A (p _A^0) = 100 torr and vapour pressure of pure liquid B (p _B^0) = 300 torr]

    A
    200 torr
    B
    140 torr
    C
    180 torr
    D
    None of these

  • Two liquids A and B from ideal solutions. At 300 K, the vapour pressure of solution containing 1 mote of A and 3 mole of B is 550 mm Hg. At the same temperature, if one more movie of B is added to this solution, the vapour pressure of the solution increasés by 10 mm Hg. Determine the vapour pressure of A and B in their pure states (in mm Hg) :

    A
    400, 600
    B
    500, 500
    C
    600, 400
    D
    None of these
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