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Weekly Quiz Competition
  • Question 1
    1 / -0
    Reaction A + B C + D is a reversible reaction; which means if we start with A + B, we get C + D, and if we start with C + D, we get A + B.
    It is given that when A + B C + D, G = (-)ve if we start with A and B.

    What will be G for the reaction C + D A + B if we start with C and D?
    Solution
    As the reaction is spontaneous from both the directions, initial change in free energy () will be negative for both the directions.
  • Question 2
    1 / -0


    Identify the product in the above reaction.
    Solution
  • Question 3
    1 / -0
    A certain weak acid has a dissociation constant 1.0 10-4. The equilibrium constant for its reaction with a strong base is:
    Solution
    HA H+ + A-
    ... (i)
    Also, HA + B+ + + H2O
    … (ii)
    By equations (i) and (ii),

  • Question 4
    1 / -0
    In the complex ion , M has four d-electrons and L is a strong field ligand. According to crystal field theory, the magnetic properties of the complex ion correspond to how many unpaired electrons?
    Solution
    In the presence of a strong field ligand, d4 configuration for octahedral complex is .
    Orbital electronic distribution can be given as follows:



    Hence, the metal ion has two unpaired electrons and it is paramagnetic in nature.
  • Question 5
    1 / -0
    CH2=CH-CH2-CH2-NH2 Product

    The major product is
    Solution
  • Question 6
    1 / -0
    The standard reduction potentials of Cu2+/Cu and Cu+/Cu are 0.339 V and 0.518 V, respectively. The standard electrode potential of Cu2+/Cu+ half-cell is
    Solution
  • Question 7
    1 / -0
    Rate constant of a reaction with a virus is 3.3 x 10-4 s-1. Time required for the virus to become 75% inactivated is: (Assume that virus growth cannot take place simultaneously.)
    Solution
    K = 3.3 10-4 s-1

    (It is clear from unit of rate constant that the given reaction is a first order reaction.)

    Now, according to the first order reaction:

    K =
    t = or 70 min
  • Question 8
    1 / -0
    One mol of the complex CoCl3.6H2O on reaction with excess of AgNO3 gives two mol of white precipitate. Thus, the complex is:
    Solution
    CoCl2.6H2O + AgNO3 2 mol white ppt. (AgCl)

    This means only two chlorine are in ionic form.

    Therefore, the given complex is:

    [Co(H2O)5Cl]Cl2.H2O
  • Question 9
    1 / -0
    If atoms of radius 'r' form face-centred cubic unit cell of edge length 'a', the correct relation is:
    Solution
    For FCC unit cell:



    Let 'r' be the radius of the sphere and 'a' be the edge length of the cube.

    As there are four spheres in FCC unit cell,

    Volume of four spheres = 4(4/3 πr3)

    In FCC, the corner spheres are in touch with the face-centred sphere. Therefore, face diagonal AD is equal to four times the radius of the sphere.

    AC = 4r

    But from the right-angled triangle ACD,

    AC =

    Or 4r =
  • Question 10
    1 / -0
    Identify 'F' in the following reaction sequence:
    Solution
  • Question 11
    1 / -0
    Which of the following is correct?
    Solution
    Specific conductance or conductivity is the measure of conductance of solution of ions present in 1 cc. Thus, by increasing dilution, the concentration decreases. Hence, the number of ions present in 1 cc decreases, resulting decrease in specific conductance.
    On the other hand, molar conductivity increases with dilution because it is the product of specific conductivity and volume. With dilution, inter-ionic interactions decrease, this further enhances molar conductivity of the solution.
  • Question 12
    1 / -0
    The weight of chlorine gas undergoing oxidation when excess of chlorine is sent into hot concentrated caustic soda solution containing 24 grams of solute in it is:

    3Cl2 + 6NaOH 5NaCl + NaClO3 + 3H2O
    Solution
    3Cl2 + 6NaOH 5NaCl + NaClO3 + 3H2O
    Number of moles of NaOH available =
    Number of moles of Cl2 that react =
    Number of moles of Cl2 oxidised =
    Mass of Cl2 gas oxidised =
  • Question 13
    1 / -0
    In a nitrosyl complex, the odd electron on nitrogen of the coordinated nitrosyl is transferred to vacant (n - 1)d subshell of the central metal. Hence, the spin-only magnetic moment of the complex K[Cr(CN)4(NH3)(NO)] assuming the metal ion to be in the strong field is:
    Solution
    In the compound K[Cr(CN)4(NH3)(NO)], Cr has +4 oxidation state; therefore, the number of unpaired electrons is 2.

    So, magnetic moment = = BM
  • Question 14
    1 / -0
    The correct stability order for the following species is:

    Solution
    Carbocations containing more number of hyper conjugation structures with resonance are more stable.
  • Question 15
    1 / -0
    Which one of the following compounds is most likely to give effervescence of CO2 with NaHCO3?
    Solution

  • Question 16
    1 / -0
    When H2O2 is added to an acidified solution of K2Cr2O7,
    Solution
    When H2O2 is added to an acidified solution of K2Cr2O7, a deep blue-violet coloured compound Cr2O(O2)2 is formed.
  • Question 17
    1 / -0
    One mole of N2, 2 moles of H2 and 3 moles of ammonia are taken in a one-litre flask. The equilibrium concentration of ammonia is: (Equilibrium constant for the decomposition of ammonia = 1.89 x 10-1 mol2 L-2)
    Solution
    N2 + 3H2 2NH3



    =

    = =
    Equilibrium constant for the formation (Kc) = = 5.26

    As Qc < Kc, concentration of NH3 increases more than 3 but less then 5 mol L-1.
  • Question 18
    1 / -0
    What happens if alkali metals are allowed to react with concentrated liquid ammonia?
    Solution
    Alkali metals dissolve in liquid ammonia to give blue solution of solvated species of the type [M(NH3)n]+. Alkali metals on dissolving in liq. ammonia fom ammoniated metal ion and ammoniated electron. Thus, paramagnetic character decreases.
  • Question 19
    1 / -0
    The major product (D) in the following reaction sequence is:

    Solution
  • Question 20
    1 / -0
    The bond energy order of and is:
    Solution
    By MOT, configuration is .

    Bond order:
    He2+; configuration is ; Bond order:
    Since the bond order of He2+ is greater than the bond order of He+.

    Thus, He2+ has higher energy than He+, because of higher stability.
  • Question 21
    1 / -0
    A solution containing NaHCO3 and Na2CO3 was treated with excess of CaCl2 solution and filtered. m1 g of precipitate was obtained. On adding NaOH drop by drop, a further m2 g was precipitated. If after addition of CaCl2, the solution was not filtered but was boiled and filtered, what would have been the total weight of the precipitate?
    Solution
    Na2CO3 + CaCl2 CaCO3 + 2NaCl
    (m1)
    2NaHCO3 + CaCl2 Ca(HCO3)2 + 2NaCl
    (Soluble)
    On addition of NaOH
    Ca(HCO3)2 + 2NaOH CaCO3 + Na2CO3 + 2H2O
    (m2)
    On heating
    Ca(HCO3)2 CaCO3 + H2O + CO2
    As in both cases, one mole of calcium bicarbonate gives one mole of precipitate; on heating, CaCO3 obtained is also m2 g.
    Thus, total weight of the precipitate = (m1 + m2) g
  • Question 22
    1 / -0
    In an organic compound X, only -NH2 groups are linked to -CH2- groups in its structure. If the ratio of mass % of NH2 and mass % of CH2 is 8 : 7, find out the ratio of mass % of C to that of H in the compound.
    Solution


    Molar mass of NH2 = 16

    Molar mass of CH2 = 14

    Thus, in the given compound, -NH2 and -CH2- units are in equal number.

    Thus,
  • Question 23
    1 / -0
    The difference in vibrational kinetic energies per mole of CO2 and SO2 at temperature T, according to law of equipartition of energy, is equal to
    Solution
    Vibrational degree of freedom per mole of CO2 (polyatomic linear) = 3n - 5 = 3 x 3 - 5 = 4

    Vibrational degree of freedom per mole of SO2 (polyatomic non-linear) = 3n - 6 = 3 x 3 - 6 = 3

    Difference in vibrational kinetic energies = = 4
  • Question 24
    1 / -0
    The relative order of basic character of the following compounds is:

    Solution

    Thus, the correct order of basic character is:
    II > V > IV > I > III
  • Question 25
    1 / -0
    A first order reaction is 50% completed in 20 minutes at 27oC and in 5 minutes at 47oC. The energy of activation of the reaction is:
    Solution
    k1(300) =

    In

    Ea =

    =

    = 55.32 kJ/mol
  • Question 26
    1 / -0
    Consider the reaction:



    If the initial concentration of A is 0.5 M, then select the correct graph.
    Solution
    Rate = Rate constant
    = [Zero order reaction]
    =
  • Question 27
    1 / -0
    Decomposition of A in a rigid closed vessel takes place as A(g) 3B(g), which follows 1st order kinetics. Reaction is started with an equimolar mixture of a gas A and an inert gas X. The initial pressure was 4 atm and at the end of 20 min, the total pressure is 6 atm. The time required for 75% reaction of A will be
    Solution
    A 3B

    At t = 0, 2 atm

    At t = 20; 2 - x + 3x

    2 - x + 3x + 2 = 6 x = 1

    Applying k = , we get

    t75% = 40 min
  • Question 28
    1 / -0
    The volume of N2 at NTP required to form a monolayer on the surface of iron catalyst is 8.15 ml per gram of the adsorbent. What will be the surface area of the adsorbent, if each nitrogen molecule occupies 16 10-20 m2?
    Solution
    Volume of N2 at STP required to cover the iron surface with monolayer,

    vm = 8.15 ml g-1

    Area occupied by single N2 molecule,

    S = 16 10-20 m2

    = 16 10-20 104 cm2 = 16 10-16 cm2

    Now, 22,400 ml of N2 at NTP contains 6.022 10-23 molecules.

    8.15 ml of N2 at NTP contains .

    Area occupied by single molecule = 16 10-16 cm2

    Area (A) occupied by 2.19 1020 molecules at NTP = 2.19 1020 16 10-16 cm2 = 35.06 104 cm2

     Surface area of the iron adsorbent = 35.06 m2
  • Question 29
    1 / -0
    Consider: FeSO4 A + B + C
    Here, (B) and (C) are gases and (A) is a red brown solid. (B) can be oxidised to (C). (B) also turns acidified K2Cr2O7 solution green. (A) dissolves in HCl to give deep yellow solution (D). (D) gives a blue coloured complex (E) with potassium ferrocyanide. Identify A, B, D and E.
    Solution

    SO2



    FeCl3 + K4[Fe(CN)6]
  • Question 30
    1 / -0
    600 ml of ozonised oxygen weighs 1 g at NTP. The volume composition of ozone and oxygen in the mixture is:
    Solution
    Volume of O3 = x

    Volume of O2 = (600 - x)

    22,400 ml of O3 = 48g
    Weight of x grams of O3 =
    22,400 ml of O2 = 32 g
    Weight of (600 - x) ml of O2 =

    Total weight of ozonised oxygen = 1 g

    Therefore,



    On solving,

    x = 200

    Volume of ozone (x) = 200 ml

    Volume of oxygen (600 - x) = 600 - 200 = 400 ml

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