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Chemistry Test - 3

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Chemistry Test - 3
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Weekly Quiz Competition
  • Question 1
    1 / -0

    If 0.5 g of a solute (molar mass 100 g mol-1) in 25 g of solvent elevates the boiling point by 1 K. The molar boiling point constant of the solvent is

    Solution

    Since molality is 1/5 and molar boiling point constant is equal to 1/molality here. Hence, correct answer is 5.

  • Question 2
    1 / -0

    Van't Hoff factor of Ca(NO3)2 is

    Solution

    The number of free ions produced in this case is 3. Hence, van't Hoff factor (i) = 3.

    Ca(NO3)2 → Ca+2 + 2NO3-1

    Van't Hoff factor (i) = 3

  • Question 3
    1 / -0

    Which of the following aqueous solutions will have the lowest vapour pressure at room temperature?

    Solution

    The change in the colligative properties depends on the number of particles of the non-volatile solute,

    The 0.1 M solution of CaCl2 will furnish 0.3 moles of ions (assuming α=1).

    Since the particles of the non- volatile solute are maximum in CaCl2 the change in the colligative properties will be maximum.

    Therefore the solution of CaCl2 will have the lowest vapour pressure.

    0.1 mol of NaCl will furnish 0.2 moles (2 x 0.1) of ions. While urea and glucose non-ionisable. Therefore, i = 1, and both will gives 0.1 moles of ions.

  • Question 4
    1 / -0

    Suppose that a sample of 100 g of water contains 0.025 g of chlorine. The concentration of chlorine in ppm will be

    Solution

  • Question 5
    1 / -0

    If 18 g of glucose is added to 178.2 g of water, then the vapour pressure of water for this aqueous solution at 100oC is

    Solution

  • Question 6
    1 / -0

    In case of a buffer solution containing equimolar amount of ammonium hydroxide and ammonium chloride, the reserve basicity of the solution is due to the presence of

    Solution

    In case of a buffer solution containing equimolar amount of ammonium hydroxide and ammonium chloride, the reserve basicity of the solution is due to the presence of ammonium hydroxide molecules.

  • Question 7
    1 / -0

    Acidic buffer solution is produced on mixing the aqueous solutions of

    Solution

    Acidic buffer solution is produced on mixing the aqueous solutions of acid and acidic salt.

  • Question 8
    1 / -0

    A 0.1 M aqueous solution of a weak acid is 2% ionised. If the ionic product of water is 1 x 10-14 mol2 dm-6, then [OH-] is

    Solution

  • Question 9
    1 / -0

    When 10-6 mole of a mono basic strong acid is dissolved in 1 litre of the solvent, its pH is

    Solution

    pH=-log[10-6]=6

  • Question 10
    1 / -0

    If the concentration of [Ag+] = 10-5 in a solution, then the concentration of [Cl-] required for the precipitation of AgCl (Ksp = 2 10-12) is

    Solution

    Use the expression: Ksp = [Ag+][Cl-]

  • Question 11
    1 / -0

    The oxidation state of sulphur in Na2S4O6 is

    Solution

    The oxidation state of sulphur in Na2S4O6 is 5/2.

  • Question 12
    1 / -0

    The amount of Zn required to produce 224 ml of H2 at STP on treatment with dilute H2SO4 will be

    Solution

  • Question 13
    1 / -0

    The oxidation potential values of A, B, C and D are -0.03 V, +0.108 V, -0.07 V and +0.1 V, respectively. The non-spontaneous cell reaction takes place between

    Solution

    Ecell = EReduction (cathode) - EReduction (anode)

    Or

    Ecell = EOxidation (anode) - EOxidation (cathode)

    A cell functions when Ecell is positive, i.e. when EReduction(cathode) > EReduction (anode) and EOxidation (anode) > EOxidation (cathode).

    Since the order of oxidation potentials is: B > D > A > C.

    Therefore, the cell will not function when A is at anode and B is at cathode.

  • Question 14
    1 / -0

    In which of the following reactions does H2O2 act as a reducing agent?

    Solution

    In the reaction Cl2 + H2O2 → 2HCl + O2, the oxidation number of oxygen changes from -1 to 0.

    Hence, H2O2 is oxidised and acts as a reducing agent.

  • Question 15
    1 / -0

    For Zn2+/Zn = - 0.76 V, Cl2/Cl- = 1.36 V, H+/ (1/2)H2 = 0 V and Fe3+/Fe2+ = 0.77 V, the increasing order of strengths as a reducing agent is

    Solution

    The substances with a lower value of reduction potential are stronger reducing agents.

    Hence, the increasing order of the strengths as reducing agents is:

    Cl2, Fe3+, H2, Zn

     

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