The formula for the determination of the density of a unit cell is:
- A$\frac{a^3 \times N_A}{Z \times M} \ g \ cm^{-3}$
- B$\frac{Z \times M}{a^3 \times N_A} \ g \ cm^{-3}$
- C$\frac{a^3 \times M}{Z \times N_A} \ g \ cm^{-3}$
- D$\frac{M \times N_A}{a^3 \times Z} \ g \ cm^{-3}$
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Similar Questions
An element has a $bcc$ structure with a cell edge of $288 \ pm$. The density of the element is $7.2 \ g \ cm^{-3}$. What is the atomic mass of the element?
Metal $M$ crystallizes into a $FCC$ lattice with the edge length of $4.0 \times 10^{-8} \ cm$. The atomic mass of the metal is $........ \ g/mol$. (Nearest integer). (Use: $N_{A} = 6.02 \times 10^{23} \ mol^{-1}$,density of metal,$d = 9.03 \ g \ cm^{-3}$)
What is the number of atoms present per unit cell of aluminium having edge length $4 \ \mathring{A}$? (Given: density of $Al = 2.7 \ g \ cm^{-3}$,atomic mass of $Al = 27 \ g \ mol^{-1}$)
$A$ 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
$(N_{A} = 6 \times 10^{23} \ mol^{-1})$
$(N_{A} = 6 \times 10^{23} \ mol^{-1})$
If a metal crystallises in a face-centred cubic $(FCC)$ structure with a metallic radius of $25 \ pm$, the number of unit cells in $1.0 \ cm^3$ of the lattice is:
DifficultAP EAMCET 2018
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