Question

An element crystallises in a b.c.c lattice with cell edge of 500 pm. The density of the element is 7.5g cm^-3. How many atoms are present in 300 g of the element?

Answer 1

a =  500 pm = 500 x 10^-10 cm

z = 2

m = 300 g

${\displaystyle Density\left(d\right)=\frac{zM}{a^2{N}_A}}$

${\displaystyle 7.5=\frac{2\times m}{{\left(500\right)}^3\times {10}^{-30}\times 6.02\times {10}^{23}}}$

${\displaystyle M=\frac{7.5\times {\left(500\right)}^3\times 10(-30)\times 6.02\times {10}^{23}}{2}}$

${\displaystyle M=282.18\text{g/mol}}$

 ${\displaystyle \text{Molar mass (M)}=\frac{\text{Mass of compound}\times N_A}{\text{Number of atoms}}}$

${\displaystyle 282.18=\frac{300\times 6.02\times {10}^{23}}{\text{Number of atoms}}}$

Number of atoms = 6.4 x 10²³