Question
Derive the relationship between molar mass, density of the substance and unit cell edge length.
✓
Answer
Relationship between the density of a substance and the edge length of the unit cell:
$1.$ If the edge length of the cubic unit cell is $‘a \ ’$, then the volume of the unit cell is $a^3.$
$2.$ Suppose that mass of one particle is $'m \ ’$ and that there are $‘n \ ’$ particles per unit cell.
$\therefore$ Mass of unit cell $= m \times n …(1)$
$3.$ The density of unit cell $(\rho ),$ which is same as density of the substance is given by :
$\rho=\frac{\text { Mass of unit cell }}{\text { Volume of unit cell }}=\frac{m \times n}{a^3}=$ Density of substance $\ \ldots \ldots (2)$
$4.$ Molar mass $(M)$ of the substance is given by:
$M =$ mass of one particle $\times$ number of particles per mole
$= m \times NA (NA$ is Avogadro number$)$
Therefore, $m=\frac{M}{N_A} \ \ldots \ldots (3)$
$5.$ Combining equations $(1)$ and $(3),$ gives
$\rho=\frac{nM}{a^3 N_{A}} \ \ldots \ldots (4)$
$1.$ If the edge length of the cubic unit cell is $‘a \ ’$, then the volume of the unit cell is $a^3.$
$2.$ Suppose that mass of one particle is $'m \ ’$ and that there are $‘n \ ’$ particles per unit cell.
$\therefore$ Mass of unit cell $= m \times n …(1)$
$3.$ The density of unit cell $(\rho ),$ which is same as density of the substance is given by :
$\rho=\frac{\text { Mass of unit cell }}{\text { Volume of unit cell }}=\frac{m \times n}{a^3}=$ Density of substance $\ \ldots \ldots (2)$
$4.$ Molar mass $(M)$ of the substance is given by:
$M =$ mass of one particle $\times$ number of particles per mole
$= m \times NA (NA$ is Avogadro number$)$
Therefore, $m=\frac{M}{N_A} \ \ldots \ldots (3)$
$5.$ Combining equations $(1)$ and $(3),$ gives
$\rho=\frac{nM}{a^3 N_{A}} \ \ldots \ldots (4)$
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