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  • I have a doubt, kindly clarify. A metai has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm-3. The molarmass of the metal is:(NA Avogadro's constant = 6.02 x 1023 mol-1)

A metai has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm-3. The molar mass of the metal is:

(NA Avogadro's constant = 6.02 x 1023 mol-1)

  • Option 1)

    20 g mol-1

  • Option 2)

    40 g mol-1

  • Option 3)

    30 g mol-1

  • Option 4)

    27 g mol-1

 

Answers (1)

As we learnt in

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

 

 

 

Density of cubic unit cell, d= \frac{ZM}{a^{3}N_{o}}

For an FCC, Z=4

Given that a=404\:\: pm

d= 2.72\: gcm^{-3}

M= \frac{da^{3}N_{o}}{Z}=\frac{2.72 \times (404 \times 10^{-10}cm)^{3} \times 6.02 \times 10^{23}}{4}=27\:gmol^{-1}

Correct option is 4.

 


Option 1)

20 g mol-1

This option is incorrect.

Option 2)

40 g mol-1

This option is incorrect.

Option 3)

30 g mol-1

This option is incorrect.

Option 4)

27 g mol-1

This option is correct.

Posted by

Vakul

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