Question

The density of $KCl$ is $1.9893gc{m}^{-3}$ and the length of a side of unit cell is $6.29092\stackrel{\circ }{A}$ as determined by X-ray diffraction. The value of Avogadro's number calculated from these data is:

• A. $6.017\times {10}^{23}$
• B. $6.023\times {10}^{23}$
• C. $6.03\times {10}^{23}$
• D. $6.017\times {10}^{19}$

Answer 1

The correct option is A $6.017\times {10}^{23}$

$KCl$ has face-centred cubic structure, i.e., $Z=4$

Avogadro's number $=\frac{Z\times M}{d\times V}$

Given that, $d=1.9893;M=74.5;V=(6.29082\times {10}^{-8}{)}^{3}c{m}^3$

Avogadro's number $=\frac{4\times 74.5}{1.9893\times (6.29082\times {10}^{-8}{)}^3}$

$=6.017\times {10}^{23}$