Question

If sodium chloride is doped with
\[1\times {{10}^{-3}}mol\] of $SrC{{l}_{2}}$ then the concentration of cation vacancies will be
A) \[1\times {{10}^{-3}}mol\]
B) \[2\times {{10}^{-3}}mol\]
C) \[3\times {{10}^{-3}}mol\]
D) \[4\times {{10}^{-3}}mol\]

Answer 1

Hint: The concentration of cation vacancies majorly depends on the number of cations formed. The number of cations formed by the dissociation of one mole of salt goes to the corresponding vacant site and gives the concentration of the vacant site of a crystal.

Complete Step by Step Answer:
Strontium chloride is dissociated into one strontium cation and two chloride ions. The dissociation can be given as follows:
$SrC{{l}_{2}}\to S{{r}^{+2}}+2C{{l}^{-}}$
Thus when sodium chloride crystal is doped with strontium chloride for each vacant site created by the sodium ion is fulfilled by the strontium ion.

Thus one mole of sodium chloride is doped with \[1\times {{10}^{-3}}mol\]or 0.001 moles of strontium chloride. The number of strontium cations formed is one. Thus the concentration of the cation vacancies will be given as \[1\times {{10}^{-3}}mol\]. So, we can write that if sodium chloride is doped with
\[1\times {{10}^{-3}}mol\] of $SrC{{l}_{2}}$ then the concentration of cation vacancies will be \[1\times {{10}^{-3}}mol\].
Thus the correct option is A.

Note: In cation vacancy one cation from the crystal is removed by some external effect thus it causes a vacancy. This vacancy is fulfilled by some nearest cation of some other crystals. Due to removal of the ion from the crystal in cation vacancy the density of the crystal decreases.