A) \[6.02\times {{10}^{18}}mo{{l}^{-1}}\]
B) \[6.02\times {{10}^{19}}mo{{l}^{-1}}\]
C) \[6.02\times {{10}^{23}}mo{{l}^{-1}}\]
D) \[6.02\times {{10}^{26}}mo{{l}^{-1}}\]
Correct Answer: A
Solution :
| [a] Doping of NaCl with \[{{10}^{-3}}\]mol% \[SrC{{l}_{2}}\] means that 100 moles of NaCl are doped with \[{{10}^{-3}}\] mol of \[SrC{{l}_{2}}\]. |
| \[\therefore \] 1 mole of NaCl is doped with \[SrC{{l}_{2}}=({{10}^{-3}}/100)\] mol \[={{10}^{-5}}\] mol |
| As each \[S{{r}^{2+}}\] ion introduces one cation vacancy, therefore concentration of cation vacancies |
| \[={{10}^{-5}}\text{mol/mol}\] of NaCl |
| \[={{10}^{-5}}\times 6.02\times {{10}^{23}}\text{mo}{{\text{l}}^{-1}}\] |
| \[=6.02\times {{10}^{18}}mo{{l}^{-1}}\] |
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