Question

An element crystallizes infcc lattice with cell edge of400pm . The density of the elements is 7gcm3 .how many atoms are present in 280g of the element?

Answer 1

hyyy mate here is your answer ⤵️

➡️
$a = 400pm = {400 \times 10}^{ - 10} cm$

ρ = 7g/cm³

mA = 280 gm.

fcc, lattice, N = 4


➡️
$d = \frac{N \: \times \: M \: }{Na \: \times {a}^{3} }$

$7 \: = \: \frac{4 \times M \: }{ {6.023 \times10 }^{23} \times ( {4 \times 10}^{ - 8} ) {}^{3} }$


M = 67.2 g/mol.

$no.of \: moles \: = \frac{mass}{molar \: mass}$

$n \: = \: \frac{280}{67.2} = \: 4.17$

$no.of \: atom = \: n \times \ \: Na$

$= \: 4.17 \times {6.023 \times 10}^{23} \:$


$= {25.11 \times 10}^{23}$


_______hope it help ✌️