
A unit cell of sodium chloride has four formula units. The edge length of the unit cell is $0.564nm$. What is the density of sodium chloride?
A) $3.89gc{m^{ - 3}}$
B) $2.16gc{m^{ - 3}}$
C) $3gc{m^{ - 3}}$
D) $1.82gc{m^{ - 3}}$
Answer
575.4k+ views
Hint: The unit cell of sodium chloride shows the fcc arrangement that means the face-centered arrangement of atoms. As fcc arrangement is present and the value of unit cell the edge length is given i.e. value of $a$ is given and the density can be calculated by given data by using a formula for density.
Complete step by step answer:
1) First of all let's analyze the given data in the question, the unit cell of sodium chloride has four formula units which mean the $Z$ effective value of sodium chloride atom is four.
2) The sodium chloride molecules edge length value is given which means $a$ the value is $0.564nm$ the value of edge length should always be taken in the angstrom unit which means we need to convert the value of nanometre to angstrom, which will be obtained by multiplying the value by $10$
3) Now let us see the density formula for sodium chloride,
$\rho = \dfrac{{{Z_{eff}} \times mol.weight}}{{{a^3} \times {N_A}}}$
Now we need to find out the value of the molecular weight of sodium chloride
Which is $NaCl = 23 + 35.5 = 58.5g$
The value of ${N_A}$ is Avogadro's number which is equal to $6.023 \times {10^{23}}$
4) Now as we got the values needed for calculation lets put the values in the density equation,
$\rho = \dfrac{{{Z_{eff}} \times mol.weight}}{{{a^3} \times {N_A}}} = \dfrac{{4 \times 58.5}}{{{{\left( {5.64 \times {{10}^{ - 8}}} \right)}^3} \times \left( {6.023 \times {{10}^{23}}} \right)}} = \dfrac{{234}}{{179.4 \times {{10}^{ - 24}} \times 6.023 \times {{10}^{23}}}} = 2.16g/c{m^3}$
Therefore, the density value of sodium chloride is calculated as $2.16g/c{m^3}$
5) As the value is shown in $2.16g/c{m^3}$.
So, the correct answer is Option B.
Note:
The sodium chloride has a unit cell shaped as cubic. The unit cell of sodium chloride looks the same if you put anions or cations on the corners in the structure. By using this data one can interpret the structure of the unit cell which helps to solve numerical on the unit cell of sodium chloride. The ${Z_{eff}}$ value is an effective nuclear charge which is equal to the number of formula units present in the unit cell.
Complete step by step answer:
1) First of all let's analyze the given data in the question, the unit cell of sodium chloride has four formula units which mean the $Z$ effective value of sodium chloride atom is four.
2) The sodium chloride molecules edge length value is given which means $a$ the value is $0.564nm$ the value of edge length should always be taken in the angstrom unit which means we need to convert the value of nanometre to angstrom, which will be obtained by multiplying the value by $10$
3) Now let us see the density formula for sodium chloride,
$\rho = \dfrac{{{Z_{eff}} \times mol.weight}}{{{a^3} \times {N_A}}}$
Now we need to find out the value of the molecular weight of sodium chloride
Which is $NaCl = 23 + 35.5 = 58.5g$
The value of ${N_A}$ is Avogadro's number which is equal to $6.023 \times {10^{23}}$
4) Now as we got the values needed for calculation lets put the values in the density equation,
$\rho = \dfrac{{{Z_{eff}} \times mol.weight}}{{{a^3} \times {N_A}}} = \dfrac{{4 \times 58.5}}{{{{\left( {5.64 \times {{10}^{ - 8}}} \right)}^3} \times \left( {6.023 \times {{10}^{23}}} \right)}} = \dfrac{{234}}{{179.4 \times {{10}^{ - 24}} \times 6.023 \times {{10}^{23}}}} = 2.16g/c{m^3}$
Therefore, the density value of sodium chloride is calculated as $2.16g/c{m^3}$
5) As the value is shown in $2.16g/c{m^3}$.
So, the correct answer is Option B.
Note:
The sodium chloride has a unit cell shaped as cubic. The unit cell of sodium chloride looks the same if you put anions or cations on the corners in the structure. By using this data one can interpret the structure of the unit cell which helps to solve numerical on the unit cell of sodium chloride. The ${Z_{eff}}$ value is an effective nuclear charge which is equal to the number of formula units present in the unit cell.
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