Elements A and B have fcc and bcc structures | vedclass

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(A) The number of atoms in $210 \ g$ of $A$ is $\frac{210}{M_A} 	imes N_A$ and in $594 \ g$ of $B$ is $\frac{594}{M_B} 	imes N_A$,where $M_A$ and $M

Vedclass · Accepted Answer

$9.9$

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(A) The number of atoms in $210 \ g$ of $A$ is $\frac{210}{M_A} 	imes N_A$ and in $594 \ g$ of $B$ is $\frac{594}{M_B} 	imes N_A$,where $M_A$ and $M_B$ are molar masses.Given: $\frac{210}{M_A} = \frac{594}{M_B} \implies \frac{M_B}{M_A} = \frac{594}{210} = 2.828$.Density formula: $d = \frac{Z 	imes M}{N_A 	imes a^3}$.For $A$ ($fcc$,$Z=4$): $7 = \frac{4 	imes M_A}{N_A 	imes a^3}$.For $B$ ($bcc$,$Z=2$): $d_B = \frac{2 	imes M_B}{N_A 	imes a^3}$.Taking the ratio: $\frac{d_B}{7} = \frac{2 	imes M_B}{4 	imes M_A} = \frac{1}{2} 	imes \frac{M_B}{M_A}$.Substituting the ratio: $\frac{d_B}{7} = \frac{1}{2} 	imes \frac{594}{210} = \frac{1}{2} 	imes 2.828 = 1.414$.$d_B = 7 	imes 1.414 = 9.9 \ g \ cm^{-3}$.

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