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Day 343: JEE Mains Chemistry

5 questions. Try each one first, then open its answer.

  1. Question 1

    Equilibrium. Asked once: 27 Jun 2022 (same type asked 4 times) See all 4 of this type

    54
     2NOCl(g)⇌2NO(g)+Cl2( g)\ 2NOCl(g) \rightleftharpoons 2NO(g) +Cl_{2}(\text{ }g) In an experiment, 2.0 moles of NOClNOCl was placed in a one-litre flask and the concentration of NONO after equilibrium established, was found to be 0.4 mol/Lmol/L. The equilibrium constant at 30∘C30^{\circ}C is ____\_\_\_\_ ×10−4\times10^{- 4}.

  2. Question 2

    Chemical Bonding and Molecular Structure. Asked once: 2021 Paper 14 (same type asked 4 times) See all 4 of this type

    34
    Number of sigma bonds in CH3CH=CH−C≡CHCH_{3}CH = CH - C \equiv CH

  3. Question 3

    Some Basic Concepts of Chemistry. Asked once: 26 Jun 2022 (same type asked 4 times) See all 4 of this type

    31
    A commercially sold conc. HClHCl is 35%HCl35\% HCl by mass. If the density of this commercial acid is 1.46 g/mLg/mL, the molarity of this solution is: (Atomic mass: Cl=35.5amu,H=1amuCl = 35.5amu,H = 1amu )

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  4. Question 4

    Electrochemistry. Asked once: 25 Jun 2022 (same type asked 4 times) See all 4 of this type

    145
    A solution of Fe2(SO4)3Fe_{2}\left( SO_{4} \right)_{3} is electrolyzed for ' xx ' min with a current of 1.5 A1.5\text{ }A to deposit 0.3482 g0.3482\text{ }g of FeFe. The value of xx is [nearest integer] Given: 1 F=96500Cmol−11\text{ }F = 96500Cmol^{- 1} Atomic mass of Fe=56 g mol−1Fe = 56\text{ }g{\text{ }mol}^{- 1}

  5. Question 5

    Coordination Compounds. Asked once: 4 Apr 2024 (same type asked 19 times) See all 19 of this type

    38
    Number of complexes from the following with even number of unpaired "d" electrons is . [V(H2O)6]3+, [Cr(H2O)6]2+, [Fe(H2O)6]3+\left\lbrack V\left( H_{2}O \right)_{6} \right\rbrack^{3 +},\ \left\lbrack Cr\left( H_{2}O \right)_{6} \right\rbrack^{2 +},\ \left\lbrack Fe\left( H_{2}O \right)_{6} \right\rbrack^{3 +}, [Ni(H2O)6]3+,[Cu(H2O)6]2+\left\lbrack Ni\left( H_{2}O \right)_{6} \right\rbrack^{3 +},\left\lbrack Cu\left( H_{2}O \right)_{6} \right\rbrack^{2 +} [Given atomic numbers: V=23,Cr=24,Fe=26V = 23,Cr = 24,Fe = 26, Ni=28,Cu=29]Ni = 28,Cu = 29\rbrack

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