If AgCl is doped with 1 times 10^{-4} mole pe | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) When $CdCl_2$ is added to $AgCl$,each $Cd^{2+}$ ion replaces two $Ag^+$ ions to maintain electrical neutrality. One $Cd^{2+}$ ion occupies the sit

Vedclass · Accepted Answer

$6.023 	imes 10^{17}$

Vedclass · Answer

(C) When $CdCl_2$ is added to $AgCl$,each $Cd^{2+}$ ion replaces two $Ag^+$ ions to maintain electrical neutrality. One $Cd^{2+}$ ion occupies the site of one $Ag^+$ ion,while the other $Ag^+$ site remains vacant. Thus,the number of cation vacancies created is equal to the number of $Cd^{2+}$ ions added. Given that $1 	imes 10^{-4}$ mole percent of $CdCl_2$ is added,this means $1 	imes 10^{-4} 	ext{ moles of } CdCl_2$ are present in $100 	ext{ moles of } AgCl$. Therefore,the number of moles of $Cd^{2+}$ ions per mole of $AgCl$ is $\frac{1 	imes 10^{-4}}{100} = 1 	imes 10^{-6} 	ext{ mol}$. The number of cation vacancies per mole is $1 	imes 10^{-6} 	imes N_A$,where $N_A = 6.023 	imes 10^{23} 	ext{ mol}^{-1}$. Number of vacancies $= 1 	imes 10^{-6} 	imes 6.023 	imes 10^{23} = 6.023 	imes 10^{17} 	ext{ mol}^{-1}$.

If $AgCl$ is doped with $1 \times 10^{-4}$ mole percent of $CdCl_2$,the number of cation vacancies (in $mol^{-1}$) is

Similar Questions

If $NaCl$ is doped with $10^{-5} \ mol$ of $AlCl_3$,what will be the concentration of cation vacancies?

Assertion : Due to Frenkel defect,there is no effect on the density of the crystalline solid.
Reason : In Frenkel defect,no cation or anion leaves the crystal.

Which of the following compounds exhibits metal deficiency defect?

In $AgBr$ crystal,the ion size lies in the order $Ag^{+} << Br^{-}$. The $AgBr$ crystal should have the following characteristics:

Write a short note on non-stoichiometric defects.

Vedclass Test Series

Exam Paper Generator

Online Exam Module