Question

An alkali metal has density $4.5g/c{m}^3.$ It has cubic unit cell with edge length $400pm$. The reaction of $7.68c{m}^3$ chunk of the metal with an excess of $HCl$ solution gives a colourless gas which occupies $4.54L$ at $0^{\circ }C$ and $1$bar. The unit cell of metal is:

• A. Simple cubic
• B. BCC
• C. FCC
• D. End-centred

Answer 1

The correct option is B BCC

$M+\mathrm{HCl}\stackrel{}{\to }\mathrm{MCl}+\frac{1}{2}{H}_2$

$1molM\equiv \frac{1}{2}mol{H}_2\therefore AgM\equiv \frac{1}{2}\times 22.7L{H}_2\therefore (7.68\times 4.5)gM\equiv \frac{1}{2}\times 22.7\times \frac{(7.68\times 4.5)}{A}L{H}_2=4.54L{H}_2\left(Given\right)\therefore A=86.4$

Now, $d=\frac{Z.M}{N_A.V}$

$4.5=\frac{Z\times 86.4}{(6\times {10}^{23})\times (400\times {10}^{-10}{)}^3}$

$\therefore Z=2$

$\Rightarrow$ Unit cell is BCC.