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 Multiple Choice QuestionsMultiple Choice Questions

361.

The property of crystalline solid is not

  • anisotropic

  • isotropic

  • hardness

  • denseness


362.

Due to Frenkel defect, the density of the ionic solids

  • does not change

  • decreases

  • increases

  • may increase or decrease


363.

The dimensions of a unit cell of a crystal are a = 0.397, b = 0.387, c = 0.504 and α = β = 90°, γ = 120° the crystal system is

  • cubic

  • hexagonal

  • orthorhombic

  • rhombohedral


364.

Sodium atom crystallizes in bcc lattice with cell edge a = 4.29 Å, the radius of sodium atom is

  • 18.6 Å

  • 1.86 Å

  • 0.186 Å

  • 37.2 Å


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365.

Which is mismatched for NaCl crystal ?

  • r+r-= 0.414 to 0.732

  • Coordination number = 6.6

  • Edge of unit cell =(r+ + r-)

  • Crystal structure = fcc


366.

The structure of ionic compound A+B identical to NaCl. If the edge lenght is 400pm and cation radius is 75pm, the radius of anion will be

  • 100 pm

  • 125 pm

  • 250 pm

  • 325 pm


367.

The length of unit cell edge of a body-centred cubic metal crystal is 352 pm. The radius of metal atom is

  • 162.4 pm

  • 152.4 pm

  • 142.4 pm

  • 156.pm


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368.

If NaCl is doped with 10-3 mol% of SrCl2, the number of cation vacancies will be

  • 6.023 × 1018

  • 1 × 10-3

  • 6 × 1012

  • 6.023 × 1023


A.

6.023 × 1018

Given, concentration of SrCl2 = 10-3 mol % concentration is in percentage so that take total 100 mL of solution

Number of moles of NaCl = 100-3 moles of SrCl2

Moles of SrCl2 is very negligible as compare to total moles so percentage always taken as 100 so that,

1 mole of NaCl is dipped with = 10-3100 mole of SrCl2

                                          = 10-5 mole of SrCl2

So, cation vacancies per mole of NaCl = 10-5 mol

Since, 1 mole = 6.023 × 1023 particles

So, cation vacancies per mole of NaCl

= 10-5 × 6.022 × 1023

= 6.022 × 1018


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369.

Match List I with List II and select the correct answer using the given codes.

List I (Crystal system) List II (Example)
A.Cubic 1. TiO2
B. Tetragonal 2.Graphite
C. Hexagonal 3. K2Cr2O7
D. Triclinic 4.ZnS

  • A    B    C    D

    2    3    4    1

  • A   B   C   D

    1   4   3    2

  • A   B   C   D

    3   2    1   4

  • A   B   C    D

    4   1    2    3


370.

The vacant space in bcc unit cell is

  • 10%

  • 23%

  • 32%

  • 46%


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