Question

an element forms bcc crystal the unit cell edge length is 260pm.calculate the density.
molar mass is 23g/mol​

Answer 1

Given:

Type of unit cell = bcc crystal

Edge length (a) = 260 pm = $\sf 260 \times 10^{-10} \ cm$

Molar mass (M) = 23 g/mol

To Find:

Density (ρ)

Answer:

Number of atoms in bcc crystal (Z) = 2

Density of unit cell:

$\boxed{ \boxed{ \bf{ \rho = \dfrac{Z \times M}{N_0 \times {a}^{3} } }}}$

$\sf N_0 \longrightarrow$ Avogadro's constant ($\sf 6.02 \times 10^{23}$

By substituting values we get:

$\rm \implies \rho = \dfrac{2 \times 23}{6.02 \times {10}^{23} \times {(260 \times {10}^{ - 10} )}^{3} } \\ \\ \rm \implies \rho = \dfrac{2 \times 23}{6.02 \times {10}^{23} \times {(2.6 \times {10}^{ - 8} )}^{3} } \\ \\ \rm \implies \rho = \dfrac{2 \times 23}{6.02 \times {10}^{23} \times 17.576 \times {10}^{ - 24} } \\ \\ \rm \implies \rho = \dfrac{2 \times 23}{105.8 \times {10}^{23 - 24} } \\ \\ \rm \implies \rho = \dfrac{ 46}{105.8 \times {10}^{ - 1} } \\ \\ \rm \implies \rho = \dfrac{ 46}{10.5 } \\ \\ \rm \implies \rho = 4.35 \: g {cm}^{ - 3}$

$\therefore$ $\boxed{\mathfrak{Density \ (\rho) = 4.35 \ g/cm^3}}$