Question

# An element 'X' (At. mass $$=40$$g $$mol^{-1}$$) having f.c.c structure, has unit cell edge length of $$400$$ pm. Calculate the density of 'X' and the number of unit cells in $$4$$g of 'X'. $$(N_A=6.022\times 10^{23}mol^{-1})$$.

Solution

## $$\text { edge length } = \text { 400 pm} = \text { 400 }\times 10^{-10} \text { cm}= \text { 4 }\times 10^{-8} \text { cm}$$$$\text { Density } = \dfrac { \text { Atomic mass} \times \text { Number of atoms in 1 unit cell } }{ \text { Avogadro's number} \times ( \text { edge length} )^3 }$$$$\text { Density } = \dfrac {40g mol^{-1} \times \text { 4 } }{ 6.022\times 10^{23}mol^{-1} \times ( \text { 4 }\times 10^{-8} \text { cm} )^3 }$$ $$\text { Density } =\text {4.15 g/cm}^3$$Volume of 4 g of 'X' $$=\dfrac { \text { 4 g }}{\text {4.15 g/cm}^3}=\text {0.964 cm}^3$$Volume of one unit cell $$=( \text { edge length} )^3= ( \text { 4 }\times 10^{-8} \text { cm} )^3=6.4 \times 10^{-23}\text { cm}^3$$The number of unit cells in 4g of 'X' $$= \dfrac {\text {0.964 cm}^3}{6.4 \times 10^{-23}\text { cm}^3}=1.5 \times 10^{22}$$Chemistry

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