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Question

A substance has bcc structure with edge length of 288 pm . the density of the element is 7.2 g cm2 . how many atoms r present in 2208g of the element ?

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Solution

ρ = ZM / a³N

Z = 2 for Bcc

ρ = 7.2

N the Avrogadro constant = 6.022x10²³

a the cubic unit cell lattice parameter = 288 ppm = 288 x 10-12 m

using formula

7.2 = 2 M / ( 288 x 10-12) 3 6.022x10²³

M = 51.9 gm

Element is Chromium

Total weight of element 2208

No of moles = 2208/ 51.9 = 42.5 moles = 42.5 x 6.022 x 1023 = 2.56 x 1025 atoms


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