Subject

Chemistry

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

11.

The correct order of ionization energies is

  • Cu > Ag > Au

  • Cu > Au > Ag

  • Au > Cu > Ag

  • Ag > Au > Cu


12.

Among the following compounds both coloured and paramagnetic one is

  • K2Cr2O7

  • VOSO4

  • (NH4)2.[TiCl2]

  • K3[Cu(CN)4]


13.

Which of the following has maximum bond energy?

  • C=C

  • C=O

  • O=O

  • N=O


14.

An organic compound crystallises in an ortho rhombic system with two molecules per unit cell. The unit cell dimensions are 12.05, 15.05 and 2.69 Å. If the density of the crystal is 1.419 g cm-3, then molar mass of compound will be 

  • 207 g mol-1

  • 209 g mol-1

  • 308 g mol-1

  • 317 g mol-1


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15.

The molarity of a solution in which 5.3 g Na2CO3 is dissolved in 500 mL will be

  • 1.0 M

  • 0.1 M

  • 0.20 M

  • 0.2 M


16.

1 mL of 0.01 N HCl is added to 999 mL solution of 0.1 N Na2SO4. The pH of the resulting solution will

  • 2

  • 7

  • 5

  • 1


17.

The equilibrium constants for the reaction,

Br2  2Br → 1

at 500 K and 700 K are 1 × 10-10 and 1 × 10-5 respectively. The reaction is

  • endothermic

  • exothermic

  • fast

  • slow


18.

A chemist wishes to prepare a buffer solution of pH = 2.90 that efficiently resists a change in pH yet contains only small concentration of buffering agents which one of the following weak acid along with its salt would be best to use 

  • m-chlorobenzoic acid (pKa = 3.98)

  • Acetoacetic acid (pKa = 3.58)

  • 2 5-dihydrobenzoic acid (pKa = 2.97)

  • p-chlorocmanic acid (pKa = 4.41)


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19.

If 5 L of H2O2 produces 50 L of O2 at NTP, H2O2 is

  • 50 volume

  • 10 volume

  • 5 volume

  • None of the above


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20.

The wave number of hydrogen atom in Lyman series is 82200 cm-1. The electron goes from

  • n3 → n2

  • n2 → n1

  • n4 → n3

  • None of these


B.

n2 → n1

According to Rydberg formula:

                     v = R1n12 - 1n22Here, R = 109677  109600 cm-1 and v = 82200 cm-1Thus,  82200 109600 = 112 - 1n22                    34 = 1 - 1n22                   1n22 = 1 - 34 = 14                     n2 = 4 The electron jumps from second orbit (n2) to ground state (n1).


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