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Density of Li crystal is 0.53 g// ...

Density of Li crystal is `0.53 g// cm^(3)`. The edge length of Li unit -call is ` 3.5Å` . Find out the number of Li atoms in a unit cell .`(N_(A)=6.0 xx 10^(23) "mol"^(-1) , M = 6.94 g "mol"^(-1))`

A

352 pm

B

527 pm

C

264 pm

D

154 pm

Text Solution

Verified by Experts

The correct Answer is:
A
Given, Li has a bcc structure
Density `(rho)=530"kg-m"^(-3)`
Atomic mass (M) = `6.94"g mol"^(-1)`
Avogadro's number `(N_(A))=6.02xx10^(23)"mol"^(-1)`
We know that number of atoms per unit cell in bcc
(Z) = 2.
`therefore` We have the formula for density,
`rho=(ZM)/(N_(A)a^(3))`
where a = edge-length of a unit cell.
`"or a"=root(3)((2xx6.94" g mol"^(-1))/(0.53"g cm"^(-3)xx06.02xx10^(23)"mol"^(-1)))`
`=root(3)(4.35xx10^(-23)"cm"^(-3))`
`=3.52xx10^(-8)cm`
`a=352"pm"`
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