Question

An ionic compound $XY$ forms a crystal structure in which ${}^{\prime }{Y}^{\prime }$ atoms form ccp lattice and ${}^{\prime }{X}^{\prime }$ atoms are so positioned that each ${}^{\prime }{X}^{\prime }$ atom touches $6$ ${}^{\prime }{Y}^{\prime }$ atoms. If the nearest distance between two ${}^{\prime }{X}^{\prime }$-atoms is $2\sqrt{2}$, then the density $\left(g/c{m}^3\right)$ of $XY$ crystal is :

(Given : At. mass of X = 14, At. Mass of Y = 50, Avogadro's number $=6\times {10}^{23}$)

• A. $5\sqrt{6}$
• B. $\frac{20}{3}$
• C. $\frac{5}{6}$
• D. $\frac{5}{3}$

Answer 1

Answer: (B)

$\frac{20}{3}$

The nearest distance between two X-atom in fcc lattice is:

$\frac{a}{\sqrt{2}}$ =$2\sqrt{2}$

and a = 4

$d=\frac{z\left(molarmass\right)}{N_A\times a^3}$

$=\frac{4\times 64}{6\times {10}^{23}\times (4\times {10}^{-8}{)}^3}$

$=\frac{40}{6}$

$=\frac{20}{3}g/c{m}^3$