
The pycnometer density of ${\text{NaCl}}$ crystal is $2.165 \times {10^3}{\text{kg}}.{{\text{m}}^{ - 3}}$ while its X-rays density is $2.178 \times {10^3}{\text{kg}}.{{\text{m}}^{ - 3}}$. The fraction of the unoccupied sites in ${\text{NaCl}}$ crystal is:
A. $5.968$
B. $5.96 \times {10^{ - 2}}$
C. $5.96 \times {10^{ - 3}}$
D. $5.96 \times {10^{ - 4}}$
Answer
481.5k+ views
Hint: Pycnometer relates to the measurement of volume and density based on Boyle’s law (volume-pressure relationship). Pycnometer measures the actual density of an object. Fractions of unoccupied crystal can be calculated from the pyknometric density and x-ray density.
Complete step by step solution:
Pyknometric density, $\rho $ is the closest estimate to actual density calculated from the crystal lattice and molecular weight of a substance. When the density of an object is calculated with the help of diffraction of X-rays by the crystal, then it is called X-ray density, ${\text{X}}$.
Measurement of density using a pycnometer is a very precise method. Pycnometric density of both solids and liquids can be calculated.
The ratio of both densities gives information about how much space is empty.
Now let’s combine all the data given above and solve the equation.
It is given that the pyknometric density, $\rho $$ = 2.165 \times {10^3}{\text{kg}}.{{\text{m}}^{ - 3}}$
X-ray density, ${\text{X}}$$ = 2.178 \times {10^3}{\text{kg}}.{{\text{m}}^{ - 3}}$
Fraction of unoccupied sites in ${\text{NaCl}}$ crystal can be calculated from the formula given below:
Fraction of unoccupied sites, ${\text{n}} = \dfrac{{{\text{X}} - \rho }}{{\text{X}}}$
Substituting the values, we get
${\text{n}} = \dfrac{{2.178 \times {{10}^{ - 3}} - 2.165 \times {{10}^{ - 3}}}}{{2.178 \times {{10}^{ - 3}}}} = \dfrac{{0.013 \times {{10}^{ - 3}}}}{{2.178 \times {{10}^{ - 3}}}} = 5.96 \times {10^{ - 3}}$
Thus the fraction of unoccupied sites in ${\text{NaCl}}$ crystal is $5.96 \times {10^{ - 3}}$
Hence, the correct option is C.
Note: The fraction of unoccupied sites means the empty spaces in the crystal. The real density is smaller than the density calculated from diffraction. Density is measured from diffraction using the volume of the unit cell. Thus all the atoms will not be present in the crystal. There would be empty spaces. From the pyknometric density, we can calculate the distance between the $Na^+$ and the $Cl^-$ ion in the ${\text{NaCl}}$ crystal.
Complete step by step solution:
Pyknometric density, $\rho $ is the closest estimate to actual density calculated from the crystal lattice and molecular weight of a substance. When the density of an object is calculated with the help of diffraction of X-rays by the crystal, then it is called X-ray density, ${\text{X}}$.
Measurement of density using a pycnometer is a very precise method. Pycnometric density of both solids and liquids can be calculated.
The ratio of both densities gives information about how much space is empty.
Now let’s combine all the data given above and solve the equation.
It is given that the pyknometric density, $\rho $$ = 2.165 \times {10^3}{\text{kg}}.{{\text{m}}^{ - 3}}$
X-ray density, ${\text{X}}$$ = 2.178 \times {10^3}{\text{kg}}.{{\text{m}}^{ - 3}}$
Fraction of unoccupied sites in ${\text{NaCl}}$ crystal can be calculated from the formula given below:
Fraction of unoccupied sites, ${\text{n}} = \dfrac{{{\text{X}} - \rho }}{{\text{X}}}$
Substituting the values, we get
${\text{n}} = \dfrac{{2.178 \times {{10}^{ - 3}} - 2.165 \times {{10}^{ - 3}}}}{{2.178 \times {{10}^{ - 3}}}} = \dfrac{{0.013 \times {{10}^{ - 3}}}}{{2.178 \times {{10}^{ - 3}}}} = 5.96 \times {10^{ - 3}}$
Thus the fraction of unoccupied sites in ${\text{NaCl}}$ crystal is $5.96 \times {10^{ - 3}}$
Hence, the correct option is C.
Note: The fraction of unoccupied sites means the empty spaces in the crystal. The real density is smaller than the density calculated from diffraction. Density is measured from diffraction using the volume of the unit cell. Thus all the atoms will not be present in the crystal. There would be empty spaces. From the pyknometric density, we can calculate the distance between the $Na^+$ and the $Cl^-$ ion in the ${\text{NaCl}}$ crystal.
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