Question

The compound
structure
Cucl has FCC
like Zns. Its density is 3.4g/cm²
What is the length of the edge of unit
cell ?​

Answer 1

Answer:

See the attachment

Explanation:

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Answer 2

Explanation:

mass of CuCl = mass of Cu + mass of Cl

. = 63.55u + 35.45u

. = 99u

Avogadro number, N = 6.022 × 10²³

Since it has fcc structure, Z = 4

Density, ρ = 3.4 g/cm³

$\frac{M \times Z}{N \times {a}^{3} } = 3.4$

$\frac{99 \times 4}{6.022 \times {10}^{23} \times {a}^{3} } = 3.4$

${a}^{3} = \frac{99 \times 4}{6.022 \times {10}^{23} \times 3.4} {cm}^{3}$

${a}^{3} = \frac{99}{6.022 \times {10}^{23} \times 0.85} {cm}^{3}$

${a}^{3} = \frac{99}{5.1187} \times {10}^{ - 23} {cm}^{3}$

${a}^{3} = 19.34 \times {10}^{ - 23} {cm}^{3}$

${a}^{3} = 193.4 \times {10}^{ - 24} {cm}^{3}$

$a = 5.78 \times {10}^{ - 8} cm$

$a = 578 \times {10}^{ - 10} cm$

$a = 578A\degree$