Question

An element has a face centered cubic unit cell with a length of 352.4pm along an edge the density of the element is 8.9 gcm-3 how many atoms present in 100 g of element

Answer 1

Answer : The number of atoms present in 100 g of element is, $10.23\times 10^{23}$

Solution : Given,

Density  = $8.9gcm^{-3}$

Number of atom in unit cell of FCC (Z) = 4

Edge length = $352.4pm=352.4\times 10^{-10}cm$ $(1pm=10^{-10}cm)$

Avogadro's number $(N_{A})=6.022\times 10^{23} mol^{-1}$

Mass of element  = 100 g

First we have to calculate the atomic mass of an element.

Formula used :  

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$      .............(1)

where,

$\rho$ = density

Z = number of atom in unit cell

M = atomic mass

$(N_{A})$ = Avogadro's number  

a = edge length of unit cell

Now put all the values in above formula (1), we get

$8.9gcm^{-3}=\frac{4\times M}{(6.022\times 10^{23}mol^{-1}) \times(352.4\times 10^{-10}Cm)^3}$

$M=58.6g/mole$

Now we have to calculate the moles of element.

$\text{Moles of element}=\frac{\text{Mass of element}}{\text{Molar mass of element}}=\frac{100g}{58.6g/mole}=1.7moles$

Now we have to calculate the number of atoms.

As, 1 mole contains $6.022\times 10^{23}$ number of atoms

So, 1.7 mole contains $1.7\times (6.022\times 10^{23})=10.23\times 10^{23}$ number of atoms

Therefore, the number of atoms present in 100 g of element is, $10.23\times 10^{23}$