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An element has body centred cubic (bcc) ...

An element has body centred cubic (bcc) structure with cell edge of 288 pm and the density of the element is `7.2g//cm^(3)`. How many atoms are present in 208 g of the element ?

A

`24.16times10^(25)`

B

`24.16times 10^(23)`

C

`24.16times10^(24)`

D

`24.16times10^(26)`

Text Solution

Verified by Experts

The correct Answer is:
B
Volume of the unit cell `=(288"pm")^(3)`
`=(288times10^(-12)m)^(3)`
`=23.9times10^(-24)cm^(3)`
Volume of 208 g of the element `=(208g)/(7.2g cm^(-3))=28.88 cm^(3)`
Number of unit cells in this volume `=(28.88)/(23.9times10^(24)) `
=`12.08times10^(23)" unit cell "`
`" Each b.c.c. cube contains 2 atoms "`
`therefore" Total number of atoms in 208 g is" `
`=2times12.08times10^(23)`
`=24.16times10^(23)" atom"`.
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