Answer
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Hint: Try to recall that density of a unit cell or crystal is given by $\rho =\dfrac{Z\times M}{{{N}_{a}}\times {{a}^{3}}}$ where, a is edge of unit cell in cm, Na= Avogadro number ($6.022\times {{10}^{23}}$ ), M= atomic mass of element or formula mass of the compound and Z=Number of atoms per unit cell or formula units. Following is the data of the number of formula mass units per cell and its name.
Complete step by step solution:
- Given, density of potassium bromide crystal, $\rho =2.75gmc{{m}^{-3}}$,
Length of the edge, a=654 pm= $654\times {{10}^{-12}}m$ = $654\times {{10}^{-10}}cm$
Now, let’s find the formula mass of KBr.
Formula mass of KBr = Atomic mass of K + Atomic mass of Br
Formula mass of KBr = 39 + 80
Formula mass of the crystal, KBr=119 g
Now, we will use the formula $\rho =\dfrac{Z\times M}{{{N}_{a}}\times {{a}^{3}}}$ and from it, we will find the value of formula mass per unit of KBr.
So, we know that $\rho =\dfrac{Z\times M}{{{N}_{a}}\times {{a}^{3}}}$ ......(1)
Now, we will put the values of $\rho ,\text{ }{{\text{N}}_{a}}$ , M and a into this equation (1).
So, we will get $2.75=\dfrac{Z\times 119}{(6.022\times {{10}^{23}})\times {{(654\times {{10}^{-10}})}^{3}}}$
Now, we can write this equation as
$Z=\dfrac{2.75\times 6.022\times {{10}^{23}}\times {{(654\times {{10}^{-10}})}^{-3}}}{119}$
$Z=0.1391\times {{10}^{23}}\times 2.79\times {{10}^{-22}}$
$Z=3.880\approx 4$
Thus, from the value obtained, we can conclude that a unit cell of Kbe contains four formula masses of KBr.
- Thus, we will take a round figure of the answer we obtained as we need a complete number.
Since, the unit cell contains 4 formula units of KBr, it is face centred cubic (FCC).
- The NaCl structure contains $C{{l}^{-1}}$ ions have CCP or FCC arrangement and $N{{a}^{+}}$ ions occupy all octahedral sites and coordination number of $C{{l}^{-1}}$ is 6 and the coordination number of $N{{a}^{+}}$ is also 6.
- Thus, we can also say that KBr crystal has similar structure as NaCl crystal.
Note: At ordinary temperatures and pressures, chlorides, bromides and iodides of Li, Na, K and Rb as well as some of the halides of silver possess the sodium chloride type structure. Also, you should remember that on application of high pressure, the NaCl structure changes to CsCl type structure in which the coordination number of ions becomes 8:8.
Number of Formula mass units per unit cell | Name of structure |
4 | FCC |
2 | BCC |
1 | Simple cubic |
Complete step by step solution:
- Given, density of potassium bromide crystal, $\rho =2.75gmc{{m}^{-3}}$,
Length of the edge, a=654 pm= $654\times {{10}^{-12}}m$ = $654\times {{10}^{-10}}cm$
Now, let’s find the formula mass of KBr.
Formula mass of KBr = Atomic mass of K + Atomic mass of Br
Formula mass of KBr = 39 + 80
Formula mass of the crystal, KBr=119 g
Now, we will use the formula $\rho =\dfrac{Z\times M}{{{N}_{a}}\times {{a}^{3}}}$ and from it, we will find the value of formula mass per unit of KBr.
So, we know that $\rho =\dfrac{Z\times M}{{{N}_{a}}\times {{a}^{3}}}$ ......(1)
Now, we will put the values of $\rho ,\text{ }{{\text{N}}_{a}}$ , M and a into this equation (1).
So, we will get $2.75=\dfrac{Z\times 119}{(6.022\times {{10}^{23}})\times {{(654\times {{10}^{-10}})}^{3}}}$
Now, we can write this equation as
$Z=\dfrac{2.75\times 6.022\times {{10}^{23}}\times {{(654\times {{10}^{-10}})}^{-3}}}{119}$
$Z=0.1391\times {{10}^{23}}\times 2.79\times {{10}^{-22}}$
$Z=3.880\approx 4$
Thus, from the value obtained, we can conclude that a unit cell of Kbe contains four formula masses of KBr.
- Thus, we will take a round figure of the answer we obtained as we need a complete number.
Since, the unit cell contains 4 formula units of KBr, it is face centred cubic (FCC).
- The NaCl structure contains $C{{l}^{-1}}$ ions have CCP or FCC arrangement and $N{{a}^{+}}$ ions occupy all octahedral sites and coordination number of $C{{l}^{-1}}$ is 6 and the coordination number of $N{{a}^{+}}$ is also 6.
- Thus, we can also say that KBr crystal has similar structure as NaCl crystal.
Note: At ordinary temperatures and pressures, chlorides, bromides and iodides of Li, Na, K and Rb as well as some of the halides of silver possess the sodium chloride type structure. Also, you should remember that on application of high pressure, the NaCl structure changes to CsCl type structure in which the coordination number of ions becomes 8:8.
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