Question

An element having the fcc lattice structure has a cell edge of 290 pm. The density of the element is $6.8g/c{m}^3$. How many atoms are present in 408 g of the element?

• A. 1.73 x 10²⁶
• B. 1.22 x 10²⁶
• C. 1.38 x 10²⁶
• D. 1.73 x 10²⁵

Answer 1

The correct option is (B)

The density of a cubic lattice can be calculated by using given formula

$d=\frac{n\times M}{a^3\times N_A}$

where $d=$ density

$n=$ number of atom present in lattice

$a=$ edge length of lattice

$M=$ mass of substance

NA=N_A=NA​= avagdaro number

For FCC lattice

$n=4$

a=290pm=290×10−10cma=290pm=290\times 10^{-10}cma=290pm=290×10−10cm

d=6.8g/cm3d=6.8g/cm^3d=6.8g/cm3

M=a3×d×NAZM=\frac{a^3\times d \times N_A}{Z}M=Za3×d×NA​​

M=6.8×(290×10−10)3×6.023×10234M=\frac{6.8\times(290\times 10^{-10})^3\times 6.023\times 10^{23}}{4}M=46.8×(290×10−10)3×6.023×1023​

$M=24.9$

Number of atoms present =6.023×1023×40824.9=\frac{6.023\times 10^{23}\times 408}{24.9}=24.96.023×1023×408​

Number of atoms present =98.6×1023=98.6\times 10^{23}=98.6×1023 atoms

$\therefore$ The number of atoms present in 408g of the element is 98.6×1024atoms98.6\times 10^{24}atoms98.6×1024atoms .

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