NEET Chemistry The Solid State Questions Solved

Atomic radius of silver is 144.5 pm. The unit cell of silver is a face centred cube. Calculate the density of silver.

(1) 10.50 g/cm3

(2) 16.50 g/cm3

(3) 12.30 g/cm3

(4) 15.50 g/cm3

(1) For (fcc) unit cell, atoms touch each other along the face diagonal.

Hence, Atomic radius (R) $=\frac{a\sqrt{2}}{4}$

$a=\text{\hspace{0.17em}}\frac{4R}{\sqrt{2}}=\frac{4}{\sqrt{2}}×144.5pm=408.70pm=408.70×{10}^{-10}cm$

Density (D) $=\text{\hspace{0.17em}}\frac{ZM}{V{N}_{0}},\text{\hspace{0.17em}}$ V = a3

D = $\frac{ZM}{{a}^{3}{N}_{0}}$; where Z for (fcc) unit cell = 4 , Avagadro’s number $\left({N}_{0}\right)=6.023×{10}^{23}$, Volume of cube (V) $={\left(408.70×{10}^{-10}\right)}^{3}c{m}^{3}$ and M (Mol. wt.) of silver = 108,

D $=\frac{4×108}{{\left(408.70×{10}^{-10}\right)}^{3}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}6.023×{10}^{23}}$$=\text{\hspace{0.17em}}10.50\text{\hspace{0.17em}}g/c{m}^{3}$

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