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Question: \[\text{AgCl}\] has the same structure as that of \[\text{NaCl}\]. The edge length of the unit cell ...

AgCl\text{AgCl} has the same structure as that of NaCl\text{NaCl}. The edge length of the unit cell of AgCl\text{AgCl}is found to be 555pm and density of AgCl\text{AgCl}is 5.561 g cm3\text{5}\text{.561 g c}{{\text{m}}^{-3}}. If the percentage of sites that unoccupied is x, the value of 100x is:

Explanation

Solution

By using the relationship between density (d), number of the atoms (z), edge length (a) and atomic mass (M) of the unit cell i.e. d = × Ma3 × NA Kg m3\text{d = }\dfrac{\text{z }\times \text{ M}}{{{a}^{3}}\text{ }\times \text{ }{{\text{N}}_{\text{A}}}}\ \text{Kg }{{\text{m}}^{-3}} , by putting all the value here, we can calculate the value of 100x.

Complete Step-by-Step Answer:
- In the given question, we have to calculate the value of 100x for which the density and edge length of silver chloride is given.
- As we know that that the relationship between density (d), number of the atoms (z), edge length (a) and atomic mass (M) of the unit cell is given by
d = × Ma3 × NA Kg m3\text{d = }\dfrac{\text{z }\times \text{ M}}{{{a}^{3}}\text{ }\times \text{ }{{\text{N}}_{\text{A}}}}\ \text{Kg }{{\text{m}}^{-3}} ….. (1)
- So, firstly, we have to calculate the molar mass of the silver chloride i.e. AgCl = 107.86 + 35.5 = 143.36 g/mol\text{AgCl = 107}\text{.86 + 35}\text{.5 = 143}\text{.36 g/mol}
- It is given that the density and edge length of the silver chloride is 5.561 g cm3\text{5}\text{.561 g c}{{\text{m}}^{-3}} and 555pm respectively.
- Also, we know that the Avogadro's number is 6.022 × 1023\text{6}\text{.022 }\times \text{ 1}{{\text{0}}^{23}}, so by putting all the value in equation (1) we will get,
d = × Ma3 × NA\text{d = }\dfrac{\text{z }\times \text{ M}}{{{a}^{3}}\text{ }\times \text{ }{{\text{N}}_{\text{A}}}} = 5.561= × 143.36(555 × 103)3 × 6.022 × 1023\text{5}\text{.561= }\dfrac{\text{z }\times \text{ 143}\text{.36}}{{{(555\text{ }\times \text{ 1}{{\text{0}}^{3-}})}^{3}}\text{ }\times \text{ 6}\text{.022 }\times \text{ 1}{{\text{0}}^{23}}}
z = 3.995\text{z = 3}\text{.995}
- Now, we know that the sodium chloride has FCC or Face Centred Cubic lattice structure so the total number of atoms will be 4.
- But in AgCl the number of atoms is 3.995.
- Thus, the percentage i.e. unoccupied by x is:
4 - 3.9954 × 100\dfrac{\text{4 - 3}\text{.995}}{4}\text{ }\times \text{ 100} = 0.12%.
- So, 100x will be equal to 100 × 0.12 = 12100\text{ }\times \text{ 0}\text{.12 = 12}

Therefore, the value of the 100x is 12.

Note: FCC or Face Centred Cubic unit cell is the arrangement of atoms in a cube in which the atoms are present at the corners (anion) and the face diagonal (cation). The contribution of each atom at the corner is 1/8 and the face diagonal it is, 1/2.