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Solutions Test - 3
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Solutions Test - 3
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  • Question 1/15
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    Only One Option Correct Type
    This section contains 11 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct
    An unopened soda has an aqueous concentration of C02 at 25 °C equal to 0.0506 molal. Thus, pressure of C02 gas in the can is (KH = 0.034 mol/kg bar) 

    Solutions

  • Question 2/15
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    CO(g)is dissolved in H2 Oat 25 °C and 0.010 atm. Henry ’s law constant for this system is 5.80 x104 atm. Thus, mole fraction of CO(g)is

    Solutions

    The mole fraction of a gas in solution is related to partial pressure of the gas above the solution by Henry ’s law.

  • Question 3/15
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     H2 S gas is used in qualitative analysis of inorganic cations. Its solubility in water at STP is 0.195 mol kg-1 . Thus, Henry ’s law constant (in atm molar-1 ) for H2 S is

    Solutions

  • Question 4/15
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     Variation of Kh (Henry ’s law constant) with temperature T is shown by following graphs I - IV.


    Correct representation is

    Solutions

    Partial pressure  

    If temperature is lowered, solubility of gases increases at a given pressure. Thus, KH increases with increase in temperature. It is due to this reason that aquatic species are more comfortable in cold water rather than in warm water.

  • Question 5/15
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    Concentration of C02 (in mole fraction) in fat when partial pressure of C02 is 55 kPa at 25 °C, is (Henry ’s law constant of C02 = 8.6 x 104 torr)

    Solutions

    Note: Kh (Henry ’s law constant) is in pressure unit, hence we use relation, Concentration xKH = Pressure

  • Question 6/15
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    What is the concentration of 02 in a fresh water stream in equilibrium with air at 25 °C and 1.0 bar? Given, KH (Henry ’s law constant) of 02 = 1.3 x 10-3 mol/kg bar at 25 °C.

    Solutions

    In this case, KH has been given in the unit of mol/kg bar. Thus, solubility of 02 (or concentration) = KH Pgas

  • Question 7/15
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    Following data has been given for C02 for the concentration in H2 0.

    If solution of C02 in H2 0 is heated from 273 to 333 K, pressure ( p2 ) needed to keep C02 in the solution is

    Solutions

  • Question 8/15
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    Relation between the volume of gas (2) that dissolves in a fixed volume of solvent(1) and the partial pressure of gas (2) is (nt = total moles, K1 and K2 are Henry ’s constants)

    Solutions

  • Question 9/15
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    Which is correct about Henry 's law

    Solutions

    Henry 's law states that at a constant temperature, the solubility of a gas is directly proportional to the pressure of a gas.
    In other words, The partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution.
    p = KH x
    KH is the henry 's law constant. 

  • Question 10/15
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    Henry ’s law constant for N2 at 293 K is 76.48 kbar. N2 exerts a partial pressure of 0.987 bar. If N2 gas is bubbled through water at 293 K, then number of millimoles of N2 that will dissolve in 1 L of water is

    Solutions

    Henry ’s law constant is in the unit of pressure, hence we use relation

    Let number of moles of nitrogen = n

    (Since, is very small)

  • Question 11/15
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    A handbook lists the solubility of carbon monoxide in water at 0 °C and 1 atm pressure as 0.0354 mL CO per mL of H2 0. What should be the pressure of CO(g) above the solution to obtain 0.010 M CO solution?

    Solutions

    Volume of CO = 0 .0354 mL = 0 .0354 x 10- 3 L

    p = 1 atm, T = 273 K

  • Question 12/15
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    One or More than One Options Correct Type
    This section contains 2  multiple type questions. Each question has 4 choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct     .

    Select the correct statement(s).

     

    Solutions

     

     

    NH3 , C02 , S03 ...... gases form ions in water, hence these gases are soluble in water. 02 , N2 .......do not form ions in water hence, insoluble thus correct.

    (b) NH3 and C02 are liquefied easily hence, soluble in water thus, correct.

    Higher the value of KH , smaller the solubility (x) of the gas at a given pressure thus, correct.

     

     

  • Question 13/15
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    Select the correct statement(s).

     

    Solutions

     

     

    (a) At constant temperature, different gases have different KH values. Thus, KH is a function of nature of gases.
    Thus, correct.
    (b) p (partial pressure)  = 
    If p is constant and KH value increases then      (solubility) decreases. Thus, correct.
    (c) KH  increases with increase in temperature thus, correct.
    (d) KH  is independent of pressure thus, incorrect.

     

     

  • Question 14/15
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    Statement Type
    This section is based on Statement I and Statement II. Select the correct anser from the codes given below
    Statement I Solubility of gases in liquids decreases with rise in temperature.
    Statement II CO2 (g) + H2 0 ( / ) ----- »H2 CO3 (aq ); ΔH = - ve

    Solutions

    (b) Dissolution of gas can be considered as the reverse of vaporisation, i.e. condensation in which heat is evolved. It is thus, exothermic ΔH <0. By Le-Chatelier ’s principle, increase in temperature shifts the dissolution equilibrium in the direction of form ation of C02 (g).
    Thus, increase in temperature decreases solubility. Thus, both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I.

  • Question 15/15
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    Statement I:  People living at high altitudes have symptoms of a condition known as anoxia.

    Statement II: At high altitudes, the partial pressure of oxygen is less than that at the ground level.

    Solutions

    At high altitudes, the partial pressure of oxygen is less than that at the ground level. This leads to the low concentration of O2 in the blood and tissues of the people residing in hill areas. This produces weakness and loss of memory.
    Symptoms of this conditions is known as ANOXIA. Thus, Statements I are correct but Statement II is the correct explanation of Statement I.

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