Question

Silver $\left(Ag\right)$ crystelise in $FCC$ lathice if edge length $\left(a\right)$ of the cell is $4.07\times {10}^{-8}cm$ & density is $10.5g/{cm}^3$. Calculate the atomic mass of $Ag$.

Answer 1

Given Data :-

Edge length of the cell $=d=4.07\times {10}^{-8}cm$

Density $=P=10.5g/{cm}^3$

(rho)

we know that

Number of atoms in a unit cell of face centered cubic $(F.C.C)$ latfice is $4$.

$\therefore Z=4$

Avogadro Number $\left(N_A\right)=6.022\times {10}^{23}$

mass of silver $=M=?$

Formula :-

Density $P=\frac{ZM}{a^3{N}_A}$

by cross multiplication, we get

$\Rightarrow {Pa}^3{N}_A=ZM$

by dividing by Z, we get

$\Rightarrow M=\frac{{Pa}^3{N}_A}{Z}$

where $P=$ density

$a=$ edge length of the cell

$N_A=$ Avogadro Number

$Z=$ Number of atoms in $F.C.C$ unit cell

$M=$ mass of the metal

by substituting the values, we get

$M=\frac{{10.5g/cm}^3\times {(4.07\times {10}^{-8}cm)}^3\times 6.022\times {10}^{23}{moles}^{-1}}{4}$

$=\frac{10.5g\times {\left(4.07\right)}^3\times {10}^{-24}{cm}^3\times 6.022\times {10}^{23}}{4{cm}^3.mole}$

$=\frac{10.5g\times {\left(4.07\right)}^3\times 6.022\times {10}^{-1}}{4mole}$

$=\frac{10.5g\times 67.41\times 6.022}{10\times 4mole}$

$=\frac{4262.401719}{40mole}$

$=106.56g/mole$

$\approx 107g/mole$