Question

The unit cell length of sodium chloride crystal is 564 pm. Its density would be:

• (A) $1.082{\mathrm{g}\mathrm{cm}}^{-3}$
• (B) $2.165{\mathrm{g}\mathrm{cm}}^{-3}$
• (C) $3.247{\mathrm{g}\mathrm{cm}}^{-3}$
• (D) $4.330{\mathrm{g}\mathrm{cm}}^{-3}$

Answer 1

The Correct Option is B

Explanation:
Given:$a=564\mathrm{pm}=564\times {10}^{-10}cm$We know that:The density is given by the formula:$\rho =\frac{zM}{a^3{N}_A}$As we know that,Each unit cell of $\mathrm{NaCl}$ has $4{\mathrm{Na}}^{-}$and $4{\mathrm{Cl}}^{-}$ions.$\Rightarrow z=4$Avogadro's number, $N_A=6.022\times {10}^{23}/\mathrm{mol}$The total mass of $\mathrm{NaCl}$:$M=22.99+34.34=58.5\mathrm{g}/\mathrm{mol}$Therefore,Density, $\rho =\frac{zM}{a^3{N}_A}$$=\frac{4\times 58.5}{{(564\times {10}^{-10})}^3\times 6.022\times {10}^{23}}{\mathrm{g}\mathrm{cm}}^{-3}$$=2.165{\mathrm{g}\mathrm{cm}}^{-3}$Hence, the correct option is (B).