Question

silver atoms are arranged in ccp lattice structure .the edge length of its unit cell is 408 pm .calculate the density of silver ​

Answer 1

Answer:

1.05 × 10∧29

Explanation:

GIVEN - ccp latice (Z) = 4

= EDGE LENGTH (a³) = 408 pm

= MASS OF SILVER (m) = 107.87

= A.T.Q = D = Z × M / Na × a³

= 4 × 107.87 / 6.022 × 10∧23 × (408)³

= ∴ DENSITY WILL BE 1.05 × 10∧29

I HOPE IT MAY HELP YOU !

Answer 2

Step by Step explanation:

Given : Silver atoms are arranged in CCP lattice structure, the edge length of its unit cell is 408 pm.

To find : The density of silver

Formula used :

$\rho = \frac{M_{Ag }*Z}{N_A *a^3}$

$\rho$ represents the density, $M_{Ag}$ is the Molar mass of silver(Ag), Z stands for no. of atoms, $N_A$ is the Avogadro number and a is the edge length of lattice.

Unit conversion :

1 pm = $10^{-10}$ cm

• Calculation for density of silver

Silver atoms are arranged in CCP (cubic close packing) or we can say FCC (Face centered cubic) so the number of atoms are;

Z = $\frac{1}{8}$ × (no. of corner atoms) + $\frac{1}{2}$ × (no. of atoms which are present on faces)

[ $\frac{1}{8}$ because corner atom contribute to 8 different cubes and $\frac{1}{2}$ is due to the atom present at face contribute to 2 cubes]

No. of corner atoms in a cube = 8   { cube has 8 corners}

No. of atoms at faces in cube = 6   {a cube has 6 faces}

$Z=\frac18*8+\frac12*6$

Z = 1 + 3

Z = 4

we have ,

edge length of unit cell (a) = 408 pm

                                            = 408 × $10^{-10$ cm

for making easy to calculate $a^3$ we can write edge length in decimals,

⇒  $a=408 *10^{-2}*10^{-8}cm$

       $=4.08 * 10^{-8}cm$

Molar mass of Silver ($M_{Ag}$ ) = 107.87 $gmol^{-1$

Avogadro number ($N_A$) = 6.023 × $10^{23$ $mol ^{-1$

and Z = 4

by substituting the data in formula of density we get the density ($\rho$) of silver,

⇒  $\rho = \frac{M_{Ag }*Z}{N_A *a^3}$

       $=\frac{107.87 gmol^{-1} * 4}{6.023 *10^{23} mol ^{-1} * {[4.08 *10^{-8}cm]}^3}$

       $=\frac{431.48g}{6.023 * 10^{23} * [67.92*10^{-24}cm^3]}$

       $=\frac{431.48g}{409.08*10^{23}*10^{-24}cm^{3}}$

       $=1.055 * 10 g cm^{-3}$

       $= 10.55 gcm^{-3$

In this way we get the density of silver is $10.55 gcm^{-3$.