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Solutions Test - 7
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Solutions Test - 7
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  • Question 1/10
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    This section contains multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct

    The vapour pressure of pure liquid solvent is 0.80 atm. When a non-volatile solute B is added to the solvent, its vapour pressure drops to 0.60 atm. Thus, mole fraction of the component B is

     

    Solutions

     

     

    Using  Raoults law for lowering of vapour pressure.We get   
    putting,Values of actual and lowered vapour pressure we get

    where  x2 ​ is the mole fraction .

     

     

  • Question 2/10
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    Vapour pressure of pure water at 298 K is 23.8 torr. Thus, vapour pressure of a solution containing 6 g urea and 68.4 g sucrose in 10 moles of water is

    Solutions

    By Raoult ’s law, for a mixture of non-volatile solute in volatile solvent.

     

  • Question 3/10
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    Vapour pressure of nitrobenzene (molar mass = 123 g mol-1 ) is 3.6 kNm-2 and that of water (molar mass = 18 g mol-1 ) is 977 kNm-2 . They form immiscible mixture at the given temperature. Thus, percentage of nitrobenzene in the vapour phase is

    Solutions

    For immiscible solvents, composition of each component is based on

     

  • Question 4/10
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    The vapour pressure of n-hexane at 350 K is 840 torr and that of cyclohexane is 600 torr. Mole fraction of hexane in the mixture that boils at 350 K and 1 atm pressure assuming ideal behaviour is

    Solutions

    d) Mixture boils at 1 atm and 350 K.
    Thus, total vapour pressure is 1 atm = 760torr (h) used for hexane and (ch) for cyclohexane.

  • Question 5/10
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    At 298 K, the vapour pressure of pure water is 23.76 mm Hg and that of an aqueous solution is 22.98 mm Hg. Thus, molality of solution is (assume dilute solution) 

    Solutions

    (c)

    By Raoult ’s law for a non-volatile solute in volatile solvent.

  • Question 6/10
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    A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the following statements is/are correct regarding the behaviour of the solution?

    Solutions

    Ethanol molecules are joined to each other by intermolecular H-bonding thus, they have high boiling point. When n-hexane is mixed into it, H-bonding breaks. Since, force of attraction between n-hexane and ethanol molecules is smaller than between the hexane-hexane and ethanol-ethanol. Thus, mixture is more volatile. This shows positive deviation from Raoult ’s law

  • Question 7/10
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    18 g of glucose (C6 H12 O6 ) is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at 100 °C is

    Solutions

     (d)

    At 100 °C, vapour pressure of water = 1 atm. Thus, p0  = 760torr Let vapour pressure of the solution = ps

     

  • Question 8/10
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    Which of the following liquid pairs shows a positive deviation from Raoult ’s law?

    Solutions

    (a) When HCI is mixed with H2 0, force of attraction between HCI and H2 O   molecules is increased (due to dissolution), hence vaporisation decreased, a case of negative deviation.

    (b) CH3 OH molecules are joined by intermolecular H-bonding which provides it a liquid state. On adding benzene, H-bonding breaks and force of attraction between CH3 OH and benzene molecules is less than between CH3 OH-CH3 OH and between benzene-benzene hence, more vaporisation of mixture, a case of positive deviation.

    (c) As in (a), negative deviation.

    (d) CH3 COCH3 and CHCI3 are volatile. On mixing, there is increase in intermolecular H-bonding due to the formation of chloretone (A)

    Thus, force of attraction between (CHCI3 -CH3 OCH3 ) molecules is larger than between (CHCI3 -CHCI3 )and (CH3 COCH3 - CH3 OC H3 ) molecules. Thus, vaporisation decreases, a case of negative deviation.

  • Question 9/10
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    A solution of two liquids boils at a temperature more than the boiling point of either of them. Hence, the binary solution shows

    Solutions

    If force of attraction between A and B (solvents) is higher than that between A and A or B and B, then there is decrease in vaporisation.
    Thus, higher temperature (i.e. boiling point) is required to convert liquids into vapour state. This is a case of negative deviation from Raoult 's law

  • Question 10/10
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    Lowering of vapour pressure of an aqueous solution of a non-volatile non-electrolyte 1 molal aqueous solution at 100 °C is

    Solutions

    (c) Molality = 1 mol kg-1  (solvent) Thus, solute = 1 mole

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