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Engineering
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At 80^{\circ}C the vapour pressure of pure liquid  A is 520 mm of Hg and that of pure liquid B is 1000  mm of  Hg. If a mixture solution of  A and B boils at 80^{\circ}C and 1 atm pressure, the amount of  A in  the mixture is \left ( 1 atm = 760 \: mm\: of Hg \right )

  • Option 1)

    50 \: mol \: percent

  • Option 2)

    52 \: mol \: percent

  • Option 3)

    34 \: mol \: percent

  • Option 4)

    48 \: mol \: percent

 
As we learnt in Rault's Law - The total vapour pressure of binary mixture of miscible liquids be having ideally is given by Where   and  are mole fraction of A and B in liquid phase.     - wherein   and are vapour pressures of pure liquids.    Given that  Let mole fraction of  & Molefraction of  at Boiling point the total pressure P is 760 mm Hg          or 50 mole percent. Correct option...
Engineering
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The vapour pressure of water at 20 ^{\circ}C is 17.5 mm  Hg  If 18 g of glucose \left ( C_{6}H_{12}O_{6} \right ) is added to 178.2 g of water at 20 ^{\circ}C the vapour pressure of the resulting solution will be

  • Option 1)

    17.325\: mm \: Hg

  • Option 2)

    17.675\: mm \: Hg

  • Option 3)

    15.750\: mm \: Hg

  • Option 4)

    16.500\: mm \: Hg

 
As we learnt in Expression of relative lowering of vapour pressure -     - wherein     Correct option is 1.   Option 1) This is the correct option. Option 2) This is an incorrect option. Option 3) This is an incorrect option. Option 4) This is an incorrect option.
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