If $M$ is the atomic mass of an element and $a$ is the edge length of the unit cell,then the formula to calculate the density $\rho$ is:
- A$\rho = \frac{n M}{a^3 N_A}$
- B$\rho = \frac{a^3 N_A}{n \times M}$
- C$\rho = \frac{a^3 M}{n N_A}$
- D$\rho = \frac{M N_A}{a^3 n}$
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Fill in the blanks:
$1.$ Density of unit cell $(d) = ........$
$2.$ Mass of atoms present in unit cell $(m) = ........$
$1.$ Density of unit cell $(d) = ........$
$2.$ Mass of atoms present in unit cell $(m) = ........$
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View SolutionHow many unit cells are present in $100 \ g$ of an element with $fcc$ crystal structure having density $10 \ g/cm^{3}$ and edge length $100 \ pm$?
$A$ metal has a $fcc$ lattice. The edge length of the unit cell is $404 \, pm$. The density of the metal is $2.72 \, g \, cm^{-3}$. The molar mass of the metal is :- ............ $g \, mol^{-1}$ ( $N_A$ Avogadro constant $= 6.02 \times 10^{23} \, mol^{-1}$ )
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View SolutionFerrous oxide has a cubic structure and each edge of the unit cell is $5.0 \ \mathring{A}$. Assuming the density of the oxide is $4.0 \ g \ cm^{-3}$,calculate the number of $Fe^{2+}$ and $O^{2-}$ ions present in each unit cell ($M_w$ of $FeO = 72 \ g/mol$).
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View Solution$KF$ has a $NaCl$ type structure. What is the distance between $K^+$ and $F^-$ in $KF$ if the density is $2.48 \ g \ cm^{-3}$?
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