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CsBr crystallises in body centred crystal lattice. The unit cell length is 436.6 pm . Given that the atomic mass of Cs is 133 and that of Br is 80 amu and Avogadro number being 6.023 \times 1023 mol-1 , the density of CsBr is

  • Option 1)

    42.5 gm/cm3

  • Option 2)

    0.425 gm/cm3

  • Option 3)

    8.25 gm/cm3

  • Option 4)

    4.25 gm/cm3

 

Answers (1)

best_answer

Number of molecule per unit cell  = 1 

Molecular mass of CsBn = 133+80=213

d=\frac{ZM}{a^{3}N_{A}}

d=\frac{1\times 213}{\left ( 4.366\times 10^{-8} \right )^{3}\times 6.023\times 10^{23}}=4.25gm/Cm^{3}

 

 

CsCl type structure -

Cl^{-} located at all corners

Cs^{+} located at body centre

Edge length = \frac{2}{\sqrt{3}} (r_{Cs^{+}} + r_{Cl^{-}})

Coordination number = 8:8

- wherein

Number of Cl^{-} = 1

Number of Cs^{+} = 1

Number of CsCl molecule per unit cell = 1

 

 

 


Option 1)

42.5 gm/cm3

This is incorrect

Option 2)

0.425 gm/cm3

This is incorrect

Option 3)

8.25 gm/cm3

This is incorrect

Option 4)

4.25 gm/cm3

This is correct

Posted by

prateek

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