Question

Edge length of FCC unit cell of copper sulphate is:

• A. $120A^o$
• B. $170A^o$
• C. $150A^o$
• D. $200A^o$

Answer 1

Answer: (C)

$150A^o$

$Z_{eff}=4$ for $CuS{O}_4$

Let the concentration be c

$Cu(H_{2}O{)}_6^{2+}+H_{2}O\rightleftharpoons [Cu\left(H_{2}O{)}_5\right(OH){]}^{+}+H_3{O}^{+}$

$c$

$c-x$ $x$ $x$

$\frac{x^2}{c-x}=\frac{{\left({10}^{-5}\right)}^2}{c-{10}^{-5}}=K={10}^{-5}$

$c=2\times {10}^{-5}$ $mol/litre$ $\Rightarrow$ moles of $CuS{o}_4=2\times {10}^{-5}\times \frac{1}{2}={10}^{-5}$ moles

Number of unit cells= $\frac{{10}^{-5}\times N_A}{4}=\frac{6}{4}\times {10}^{18}=1.5\times {10}^{18}$

Number of unit cells along one edge of the cube=$\sqrt[3]{31.5\times {10}^{18}}=1.14\times {10}^6$

$⟶$ Let the edge length of FCC be a

$a\times 1.14\times {10}^6=17.1mm\Rightarrow a=150\mathring{A}$