Question

What is the volume of a face centred cubic unit
cell, when its density is 2.0 g cm and the molar
mass of the substance is 60.22 g mol"?
(1) 4 x 10-22 cm cube
(2) 2x 10-22 cm cube
(3) 44 x 10-22 cm cube
(4) 22 x 10-22 cm cube​

Answer 1

Answer:

Volume of a Face Centred Cubic (FCC) Unit Cell is 2 x 10⁻²² cm³

Explanation:

Given ::

1. It is a FCC Unit Cell.

2. Density => 2.0 g cm⁻³

3. Mass of substance => 60.22 g mol⁻¹

Now, we know that for FCC ,

Z = 4 and Z is the no. of atoms present per unit cell or formula units.

Molar Mass = 60.23 gram

Volume of unit cell => a³ = ( Molar Mass * Z ) / ( Density * Nₐ )

a³ = ( 60.23 * 4 ) / ( 2 * 6.022 * 10²³ )

a³ = ( 60.23 * 2 ) / ( 6.022 * 10²³ )

a³ = ( 10.001 * 2 ) / 10²³

a³ = 20.001 * 10⁻²³ cm³

a³ = 2 x 10⁻²² cm³

Here,

Nₐ = Avogadro Number = 6.022 * 10²³

Answer 2

Solution:

Given:

=> It is a FCC structure , so the value of Z = 4.

=> Molar mass = 60.23 g

=> Density = 2 g/cm³

To Find:

=> Volume of unit cell (a³)

Formula used:

$\sf{\implies Volume\;of\;unit\;cell,\;a^{3}=\dfrac{Molar\;mass \times Z}{Density\times N_{A}}}$

Here Nₐ = Avogadro Number

              = 6.022 × 10²³

So,

$\sf{\implies Volume\;of\;unit\;cell,\;a^{3}=\dfrac{Molar\;mass \times Z}{Density\times N_{A}}}$

Put the value in the formula,

$\sf{\implies Volume\;of\;unit\;cell,\;a^{3}=\dfrac{60.22 \times 4}{2\times 6.022\times 10^{23}}}$

$\sf{\implies 20.0033\times 10^{-23}\;cm^{3}}$

$\large{\boxed{\boxed{\sf{Hence,\;Volume\;of\;unit\;cell,\;a^{3}=20.0033\times 10^{-23}\;cm^{3}}}}}$