Question

The density of mercury is $13.6g/ml$. The approximate diameter of an atom of mercury assuming that each atom is occupying a cube of edge length equal to the diameter of mercury atom is $a=2.9004\times {{10}^{-x}}cm$. The value of x is _____. (Given atomic mass of mercury is $200.59u$)

Answer 1

Hint: The structure is cubic and we have to find out its volume ($V=\dfrac{4}{3}\pi {{r}^{3}}$). Then we have to apply the density formula to find the unknown variable ($Density=\dfrac{Mass}{Volume}$ ).

Complete step by step solution:
Given data:
The density of mercury (D) is given as $13.6g/ml$. The diameter of a mercury atom is$a=2.9004\times {{10}^{-x}}cm$which is also equal to the edge length of the cube inside which this atom is placed. The shape is somewhat like this:

We also know the mass of a mercury atom. It is $200.59u$ where “u” is the atomic mass unit. Converting it into grams it becomes
\[200.59u\times 1.66\times {{10}^{-24}}=3.329\times {{10}^{-22}}g\]
The volume (V) of atom is $\dfrac{4}{3}\pi {{r}^{3}}$
\[\begin{align}
  & \Rightarrow V=\dfrac{4}{3}\times 3.14\times {{(1.4502\times {{10}^{-X}})}^{3}} \\
 & \Rightarrow V=12.768\times {{10}^{-3X}}c{{m}^{3}} \\
\end{align}\]
Now we have the values of all the variables required for this question. Below we put them together systematically and try to solve them. We start with the formula for density:
\[\begin{align}
  & Density(D)=\dfrac{Mass(M)}{Volume(V)}=\dfrac{3.329\times {{10}^{-22}}g}{12.768\times {{10}^{-3x}}c{{m}^{3}}} \\
 & D=0.26\times {{10}^{-22+3x}}g/c{{m}^{3}} \\
 & \text{As we know density is given as D=13}\text{.6g/ml or 13}\text{.6c}{{\text{m}}^{3}} \\
 & \text{So the above equation becomes:} \\
 & \Rightarrow 13.6=0.26\times {{10}^{-22+3x}} \\
 & \Rightarrow {{10}^{-22+3x}}=\dfrac{13.6}{0.26}=52.3 \\
 & \text{Taking the log of both sides we get:} \\
 & \Rightarrow (3x-22)log(10)=log(52.3) \\
 & \Rightarrow 3x-22=1.718 \\
 & \Rightarrow 3x=23.718 \\
 & \Rightarrow x=7.906\approx 8 \\
\end{align}\]
The value of the unknown x = 8

So, the diameter of the atom or the edge of the cube $a=2.9004\times {{10}^{-8}}cm$.

Note: The question gives the data that the edge length of the cube is the same as the diameter of the sphere. If here a student takes the volume of the cube rather than that of the sphere it would be wrong. You have to remember that the density given is that of mercury, which means it relates to the atoms of the element and not the space they occupy at the atomic level.
As you can see in the picture represented at the start, the volume of the cube with the given value of edge length is more than that of the sphere.