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Weekly Quiz Competition
  • Question 1
    4 / -1

    During dissolution when solute is added to the solvent, some solute particles separate out from the solution as a result of crystallisation. At the stage of equilibrium, the concentration of solute in the solution at given temperature and pressure

    Solution

    At dynamic equilibrium, number of solute particles going into the solution will be equal to solute particles separating out. Hence, the concentration of solute in the solution remains constant.

  • Question 2
    4 / -1

    Consider the two figures given below.

    Which of the following statements regarding the experiment is true?

    Solution

    The solubility of gas in a liquid increases with increase in pressure and is directly proportional to the pressure of the gas.

  • Question 3
    4 / -1

    The law which indicates the relationship between solubility of a gas in liquid and pressure is________.

    Solution

    Henry's law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of the liquid or solution.
    p = KHX

  • Question 4
    4 / -1

    According to Henry's law the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution. For different gases the correct statement about Henry's constant is

    Solution

    p = KHx. Higher the value of KH at a given pressure, lower is the solubility of the gas in the liquid.

  • Question 5
    4 / -1

    The value of Henry's law constant for some gases at 293K is given below. Arrange the gases in the increasing order of their solubility.
    He = 144.97kbar; H2 ​= 69.16kbar
    N2 ​= 76.48kbar; O2 ​= 34.86kbar

    Solution

    Higher the value of KH, lower is the solubility of gas in the liquid.

  • Question 6
    4 / -1

    H2S is a toxic gas used in qualitative analysis. If solubility of H2S in water at STPSTP is 0.195m, what is the value of KH?

    Solution

    No. of moles of H2S = 0.195
    No. of moles of H2O = 1000/18 = 55.55mol
    Mole fraction of H2S =
    Pressure at STP = 0.987 bar
    According to Henry’s law, p = KHx
    or KH = pH2S/xH2S = 0.98/70.0035
    = 282 bar

  • Question 7
    4 / -1

    Henry's law constant for the molality of methane in benzene at 298 K is 4.27 × 105 mm Hg. The mole fraction of methane in benzene at 298 K under 760 mm Hg is

    Solution

    According to Henry's law, p = KHx
    x = p/KH
    = 1.78 x 10-3

  • Question 8
    4 / -1

    When a gas is bubbled through water at 298K, a very dilute solution of gas is obtained. Henry’s law constant for the gas is 100kbar. If gas exerts a pressure of 1bar1bar, the number of moles of gas dissolved in 11 litre of water is

    Solution

    p = KH × x

    Mole fraction = Moles of gas/Total moles
    Moles of H2O = 1000/18 = 55.55 (∵ 1L = 1000 g)
    Mole fraction = (55.55 >>> x)

  • Question 9
    4 / -1

    How much oxygen is dissolved in 100mL water at 298K if partial pressure of oxygen is 0.5atm and KH = 1.4 × 10−3 mol / L / atm?

    Solution

    According to Henry’s law, s = KH × p, where s is concentration of O2 dissolved.
    s = 1.4 × 10−3 × 0.5 = 0.7 × 10−3mol/L
    s = n/V or n = 0.7×10−3 × 0.1 = 0.7 × 10−4mol
    n = w/M or w = n × M
    = 0.7 × 10−4 × 32 = 22.4 × 10−4g or 2.24mg

  • Question 10
    4 / -1

    At high altitudes the partial pressure of oxygen is less than that at the ground level. This leads to

    Solution

    At high altitudes the partial pressure of oxygen is less than that at the ground level. This leads to low concentrations of oxygen in the blood and tissues of people living at high altitudes or climbers.

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