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Question

The edge length of the unit cell of NaCl crystal lattice is 5.623˚A, density is 2.16 g cm3 and the molar mass of NaCl is 58.5 g mol1, the number of moles per unit cell is :

(NA is Avagadro number)

A
4NA
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B
3NA
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C
1NA
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D
2NA
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Solution

The correct option is B 4NA
The expression for density is given below:

d=zMNAa3 =z×58.56×1023×(5.623×108)3=2.16 g cm3

z=4

So, the number of pairs of NaCl per unit cell is 4.
We know that 1 mole = NA no. of ions
4 NaCl ions =4NA moles
Hence, the total no. of moles present per unit cell = 4NA moles

Option A is correct.

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