Question

Edge length (in $\mathring{A}$) of FCC unit cell of copper sulphate is :

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

Answer 1

The correct option is C $150$

The balanced reaction is as follows:

$Cu(H_{2}O{)}_6^{2+}+H_{2}O\rightleftharpoons [Cu\left(H_{2}O{)}_5\right(OH){]}^{+}+H_3{O}^{+}$
$C$ (Initial)
$C-x$ $x$ $x$ (Final)
${10}^{-5}=\frac{x^2}{C-x}=\frac{({10}^{-5}{)}^2}{C-{10}^{-5}}$ $⟹C=2\times {10}^{-5}$ mol/L
Moles of $CuS{O}_4$ dissolved $=2\times {10}^{-5}\times 0.5={10}^{-5}$
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 $=$ $\sqrt[3]{31.5\times {10}^{18}}=1.14\times {10}^6$
If edge length of FCC unit cell is $a$, then
$\therefore a\times 1.14\times {10}^6=17.1\Rightarrow a=150\mathring{A}$