A substance has a density of 2 g cm^{-3}. I | vedclass

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(B) The density of a unit cell is given by the formula: $d = \frac{Z 	imes M}{a^3 	imes N_A}$ For an $fcc$ crystal,the number of atoms per unit cell

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$64.8$

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(B) The density of a unit cell is given by the formula: $d = \frac{Z 	imes M}{a^3 	imes N_A}$ For an $fcc$ crystal,the number of atoms per unit cell,$Z = 4$. Given: $d = 2 \ g \ cm^{-3}$,$a = 600 \ pm = 600 	imes 10^{-10} \ cm = 6 	imes 10^{-8} \ cm$,$N_A = 6 	imes 10^{23} \ mol^{-1}$. Substituting the values: $2 = \frac{4 	imes M}{(6 	imes 10^{-8})^3 	imes 6 	imes 10^{23}}$ $2 = \frac{4 	imes M}{216 	imes 10^{-24} 	imes 6 	imes 10^{23}}$ $2 = \frac{4 	imes M}{216 	imes 0.1} = \frac{4 	imes M}{21.6}$ $M = \frac{2 	imes 21.6}{4} = 10.8 	imes 1 = 10.8$ Wait,recalculating: $2 = \frac{4 	imes M}{216 	imes 10^{-24} 	imes 6 	imes 10^{23}} = \frac{4 	imes M}{1296 	imes 10^{-1}} = \frac{4 	imes M}{129.6}$ $M = \frac{2 	imes 129.6}{4} = 64.8 \ g \ mol^{-1}$.

$A$ substance has a density of $2 \ g \ cm^{-3}$. It crystallizes in the $fcc$ crystal with an edge length of $600 \ pm$. The molar mass of the substance (in $g \ mol^{-1}$) is
$(N_{A} = 6 \times 10^{23} \ mol^{-1})$

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$A$ substance has a density of $2 \ g \ cm^{-3}$. It crystallizes in the $fcc$ crystal with an edge length of $600 \ pm$. The molar mass of the substance (in $g \ mol^{-1}$) is $(N_{A} = 6 \times 10^{23} \ mol^{-1})$