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
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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.
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