Question

The density of mercury is $13.6gm{L}^{-1}$. The approximate diameter of an atom of mercury assuming that each atom is occupying a cube of edge length equal to the diameter of the mercury atom is $a=2.9004\times {10}^{-x}cm$. The value of $x$ is _____.

Answer 1

Let the diameter of one mercury atom be $a$.

Volume of mercury molecule = $a^3$ (volume of cube)

Volume of $N$ mercury molecules = $a^3\times N$

$Density=\frac{Mass}{Volume}=\frac{200}{a^3\times N}=13.6$

$a^3=\frac{200}{13.6\times N}=\frac{200}{13.6\times 6.023\times {10}^{23}}$

or $a=\sqrt[3]{3\frac{200}{13.6\times 6.023\times {10}^{23}}}$

$=2.9004\times {10}^{-8}cm$

So the value of $x$ is 8.