The density of Li metal is 0.53 g/cm^3 and t | vedclass

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(A) The density formula for a unit cell is given by: $d = \frac{n 	imes M}{a^3 	imes N_A}$Given values:$d = 0.53 \ g/cm^3$$a = 3.5 \ \mathop A\limit

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$2$

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(A) The density formula for a unit cell is given by: $d = \frac{n 	imes M}{a^3 	imes N_A}$Given values:$d = 0.53 \ g/cm^3$$a = 3.5 \ \mathop A\limits^o = 3.5 	imes 10^{-8} \ cm$$M = 6.94 \ g \ mol^{-1}$$N_A = 6.02 	imes 10^{23} \ mol^{-1}$Rearranging the formula to find $n$:$n = \frac{d 	imes a^3 	imes N_A}{M}$Substituting the values:$n = \frac{0.53 	imes (3.5 	imes 10^{-8})^3 	imes 6.02 	imes 10^{23}}{6.94}$$n = \frac{0.53 	imes 42.875 	imes 10^{-24} 	imes 6.02 	imes 10^{23}}{6.94}$$n = \frac{13.67}{6.94} \approx 1.97 \approx 2$Thus,the number of $Li$ atoms per unit cell is $2$.

The density of $Li$ metal is $0.53 \ g/cm^3$ and the edge length of its unit cell is $3.5 \ \mathop A\limits^o$. Calculate the number of $Li$ atoms in the unit cell. $(N_A = 6.02 \times 10^{23} \ mol^{-1}, M = 6.94 \ g \ mol^{-1})$

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