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IfAl^(3+)replacesNa^(+)at the edge centre of NaCl lattice then calculate that vacanicles in 1 mole NaCl. |
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Answer» SOLUTION :1 mole of NaCl contains1 mole of ` Na^(+)`IONS. ` 6.023 xx 10^(23)Na^(+)` ions . NaCl has fcc arragement of `Cl^(-)` ions and ` Na^(+)`ions are present at eh edge CENTRES and body - centre. As there are 12 edge centre and each edge centre is shared by 4 shared by 4 unitcells, their contribution per unit cell ` = 1/4 xx 12 = 3` Conribution of ` Na^(+) ` ionat the body -centre =1 Thus, for every ` 4 Na^(+)`ions, the ions present at the edge centres = 3 . This means that `N^(+)`ions which havebeen replaced ` = 3/4 xx 6.023 xx 10^(23)=4.517 xx 10^(23)` ` 1Al^(3+)`ion will REPLACE ` 3Na^(+)`ions to maintain electrical neutrality . one vacancy will be occupied by ` Al^(3+)` ion and the remaining 2 will be vacent. The means that ` 1/3` rd of these positions will be occupied by `Al^(3+)`ions and ` 2/3` rd will remain vacant. Hence, no. of vanancies in1 mole of ` NaCl = 2/3 xx 4.517 xx 10^(23)= 3.01 xx 10^(23)` |
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