Question

The density of solid argon is 1.65 g per cc at -233\[^{o}C\]. If the argon atom is assumed to be a sphere of radius \[1.54\times {{10}^{-8}}\]cm, the percentage of empty space in solid argon is:
(A )32%
(B) 54%
(C) 68%
(D) 62%

Answer 1

Hint: The volume of one atom of Argon can be calculated by using the following formula.
Volume of one atom of Ar = \[\dfrac{4}{3}\pi {{r}^{3}}\]
Where = radius of the sphere or radius of the atom.

Complete step by step answer:
-In the question it is given that the density of solid argon is 1.65 g per cc = 1.65 g/ml.
-Means 1.65 g of solid argon has 1 ml volume.
-We know that volume of one atom of Argon = \[\dfrac{4}{3}\pi {{r}^{3}}\to (1)\]
-We have to calculate the number of atoms in 1.65 g or in 1 ml of solid Argon
-To get the number of atoms we have to multiply the number of moles of argon with Avogadro number is as follows.
Number of atoms in 1.65 g of solid argon
\[\begin{align}
  & =\left( \dfrac{1.65}{40}mol \right)\left( 6.023\times {{10}^{23}} \right) \\
 & =\dfrac{1.65\times 60.23\times {{10}^{23}}}{40} \\
\end{align}\]
-Now we have to substitute the number of atoms in 1.65 g of solid argon in equation (1) to get the total volume of all Argon atoms in 1 ml of Argon.
-Total volume of all Argon atoms in 1 ml of Argon = \[\begin{align}
& =\dfrac{4}{3}\pi {{r}^{3}} \\
& =\dfrac{4}{3}\times \dfrac{22}{7}\times {{\left( 1.54\times {{10}^{-8}} \right)}^{3}}\times \dfrac{1.65\times 60.23\times {{10}^{23}}}{40} \\
 & =0.380ml\text{ }or\text{ }0.380c{{m}^{3}} \\
\end{align}\]
-Given volume of solid Argon = 1 ml
-Therefore empty space = (1-0.380) ml = 0.620 ml.
-Therefore the percentage of empty space in empty Argon \[=\dfrac{0.620}{1}\times 100=62%\]
-The percentage of empty space present in solid Argon is 62%.

-So, the correct option is D.

Note: Don’t be confused with g per cc or g/ml. Both are the same.
We can measure the volume of any gas or solid by using g per cc or g/ml.
1 g per cc = 1 g/ml.