Question

In a NaCl type crystal distance between $N{a}^{+}$ and $C{l}^{-}$ ion is $2.814\mathring{A}$ and the density of solid is $2.167gm/c{m}^3$ then find out the value of Avogadro constant.

• A. $6.05\times {10}^{23}$
• B. $3.02\times {10}^{23}$
• C. $12.10\times {10}^{23}$
• D. $22.4\times {10}^{23}$

Answer 1

The correct option is A $6.05\times {10}^{23}$

$dofNaCl=2.167g/c{m}^3$

$ZofNaCl=4$

Because F.C.C. structure $r_{N{a}^{+}}+R_{C{l}^{-}}=a/2$

$281.4=0.3535a\Rightarrow a=794.9pm=794.9\times {10}^{-10}cm$

$mofNaCl=23+35.5=58.5$

$d=\frac{n\times m}{a^3\times N_A}$

$2.167=\frac{4\times 58.5}{(794.9\times {10}^{-10}{)}^3\times N_A}$

$N_A=\frac{4\times 78.5}{(794.9\times {10}^{-10}{)}^3\times 2.165}$

$N_A=6.09\times {10}^{23}$

Hence, the value of the Avogadro's number is $6.09\times {10}^{23}$.