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An element has a body centred cubic structure with a edge length of 444 pm. The density of the element is 7.2 gm/cm3 . How many atoms are present in 404 gm of element ?

  • Option 1)

    3.12 \times 1024

  • Option 2)

    9.24 \times 1024

  • Option 3)

    12.82 \times 1023

  • Option 4)

    6.41 \times 1023

 

Answers (1)

Volume of the unit cell =\left ( 444\times 10^{-10} \right )^{3}

                                   =8.75\times 10^{-23}Cm^{3}

Density=\frac{Mass}{Volume}

=\frac{Z\times Mass\; of\; one\; atom}{volume\; of\; unit\; cell}

7.2=\frac{2\times 404}{8.75\times 10^{-23}\times N}

N=\frac{808}{7.2\times 8.75\times 10^{-23}}=12.82\times 10^{23}

 

Density of cubic unit cell -

d = \frac{zM}{a^3N_o}

 

- wherein

Where,

d = density of crystal

z = no. of effective constituent particles in one unit cell

M = molecular weight

a = edge length of unit cell

No = 6.022*1023

 

 


Option 1)

3.12 \times 1024

This is incorrect

Option 2)

9.24 \times 1024

This is incorrect

Option 3)

12.82 \times 1023

This is correct

Option 4)

6.41 \times 1023

This is incorrect

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subam

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