An element with molar mass 2.7 times 10^{-2} | vedclass

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(D) The density formula is $d = \frac{z 	imes M}{N_A 	imes a^3}$.Given: $M = 2.7 	imes 10^{-2} \ kg \ mol^{-1}$,$a = 405 	imes 10^{-12} \ m$,$d =

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$143$

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(D) The density formula is $d = \frac{z 	imes M}{N_A 	imes a^3}$.Given: $M = 2.7 	imes 10^{-2} \ kg \ mol^{-1}$,$a = 405 	imes 10^{-12} \ m$,$d = 2.7 	imes 10^3 \ kg \ m^{-3}$,$N_A = 6.022 	imes 10^{23} \ mol^{-1}$.Substituting the values: $2.7 	imes 10^3 = \frac{z 	imes 2.7 	imes 10^{-2}}{6.022 	imes 10^{23} 	imes (405 	imes 10^{-12})^3}$.Solving for $z$: $z = \frac{2.7 	imes 10^3 	imes 6.022 	imes 10^{23} 	imes 6.643 	imes 10^{-29}}{2.7 	imes 10^{-2}} \approx 4$.Since $z = 4$,the unit cell is face-centered cubic $(fcc)$.For $fcc$,the relation between edge length $a$ and radius $r$ is $a = 2\sqrt{2}r$,so $r = \frac{a}{2\sqrt{2}}$.$r = \frac{405 	imes 10^{-12}}{2 	imes 1.414} = \frac{405 	imes 10^{-12}}{2.828} \approx 143.2 	imes 10^{-12} \ m$.Rounding to the nearest integer,$r = 143 	imes 10^{-12} \ m$.

An element with molar mass $2.7 \times 10^{-2} \ kg \ mol^{-1}$ forms a cubic unit cell with edge length $405 \ pm$. If its density is $2.7 \times 10^{3} \ kg \ m^{-3},$ the radius of the element is approximately......... $\times 10^{-12} \ m$ (to the nearest integer).

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