
Estimate the average mass density of a sodium atom assuming its size to be about $2.5$. (Use the known values of Avogadro's number and the atomic mass of sodium). Compare it with the density of sodium in its crystalline phase $970 Kg m^{-3}$. Are the two densities of the same order of magnitude? If so, why?
Answer
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Hint: We have to compare the densities of sodium atoms in two phases: solid and crystalline. For solid state, mass has given and the diameter is given. First, we will find the density of the sodium in solid state as density us the ratio of mass to volume. Keeping the powers of both densities same, compare the both densities.
Complete answer:
Given:
The diameter of the sodium atom is = $2.5 A$
The volume of the sodium atom is given by: -
$V = \dfrac{4}{3} \pi R^{3}$
$V = \dfrac{4}{3} \pi (1.25 \times 10^{-10})^{3} m^{3}$
Atoms in 1 mole of sodium =$6.023 \times 10^{23}$
The mass of 1 mole of sodium is $23 g = 23 \times 10^{-3} Kg$
The mass of one sodium atom is:
$m = \dfrac{23 \times 10^{-3}}{6.023 \times 10^{23}}$ Kg
Density of the atom is:
$\rho = \dfrac{m}{V}$
The density of the sodium atom is:
$\rho = \dfrac{\dfrac{23 \times 10^{-3}}{6.023 \times 10^{23}} Kg}{\dfrac{4}{3} \pi (1.25 \times 10^{-10})^{3} m^{3}}$
$\rho = 4.692 \times 10^{-3 } Kg m^{-3}$
Given that the density of sodium in the crystalline phase is $970 Kg m^{-3}$.
Therefore, the densities of sodium in two different phases are not of the same order, but in different order. In the solid phase, atoms are tightly packed. Thus, the inter-atomic distance is very small in the crystalline phase.
Additional Information:
In solid phase, particles are bound to fixed position. They can vibrate about their mean position. In the crystalline phase, atoms are regularly placed and repetitive. A change of state from one to another is called a phase change. It is associated with the change in energy.
Note:
Density is defined as ratio of mass to volume. Mass should be in kilogram; volume should be in $m^{3}$. Mass of one atom should be divided by volume of one atom. Mass of one mole divided by Avogadro’s number gives the mass of one atom. Radius should be in meters.
Complete answer:
Given:
The diameter of the sodium atom is = $2.5 A$
The volume of the sodium atom is given by: -
$V = \dfrac{4}{3} \pi R^{3}$
$V = \dfrac{4}{3} \pi (1.25 \times 10^{-10})^{3} m^{3}$
Atoms in 1 mole of sodium =$6.023 \times 10^{23}$
The mass of 1 mole of sodium is $23 g = 23 \times 10^{-3} Kg$
The mass of one sodium atom is:
$m = \dfrac{23 \times 10^{-3}}{6.023 \times 10^{23}}$ Kg
Density of the atom is:
$\rho = \dfrac{m}{V}$
The density of the sodium atom is:
$\rho = \dfrac{\dfrac{23 \times 10^{-3}}{6.023 \times 10^{23}} Kg}{\dfrac{4}{3} \pi (1.25 \times 10^{-10})^{3} m^{3}}$
$\rho = 4.692 \times 10^{-3 } Kg m^{-3}$
Given that the density of sodium in the crystalline phase is $970 Kg m^{-3}$.
Therefore, the densities of sodium in two different phases are not of the same order, but in different order. In the solid phase, atoms are tightly packed. Thus, the inter-atomic distance is very small in the crystalline phase.
Additional Information:
In solid phase, particles are bound to fixed position. They can vibrate about their mean position. In the crystalline phase, atoms are regularly placed and repetitive. A change of state from one to another is called a phase change. It is associated with the change in energy.
Note:
Density is defined as ratio of mass to volume. Mass should be in kilogram; volume should be in $m^{3}$. Mass of one atom should be divided by volume of one atom. Mass of one mole divided by Avogadro’s number gives the mass of one atom. Radius should be in meters.
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