Density of Li atom is ` 0.53 " g/cm"^(3)`, the edge length of Li is 3.5 Å . Find out the number of Li atoms in a unit cell ` ( N_(0) = 6.023 xx 10^(23) , M = 6.94)

Density of Li atom is ` 0.53 " g/cm"^(3)`, the edge length of Li is 3.

1.	Density of Li atom is ` 0.53 " g/cm"^(3)`, the edge length of Li is 3.5 Å . Find out the number of Li atoms in a unit cell ` ( N_(0) = 6.023 xx 10^(23) , M = 6.94)
Answer» The aim is to find Z in the formula ` p = ( Z xx M)/ (a^(3) xx N_(0)) ` ` Z= ( p xx a^(3) xx N_(0)) /M = ( 0.53 g cm ^(-3) xx ( 3.5 xx 10^(-8) cm)^(3) xx ( 6.023 xx 10^(3) mol^(-1)))/(6.94 " g mol"^(-1)) = 1.97 = 2 `

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Density of Li atom is ` 0.53 " g/cm"^(3)`, the edge length of Li is 3.5 Å . Find out the number of Li atoms in a unit cell ` ( N_(0) = 6.023 xx 10^(23) , M = 6.94)

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