Solveeit Logo
Login

Question

Question: If \(NaCl\) is doped with \(10^{- 3}mol\mspace{6mu}\% SrCl_{2},\) then the concentration of cation ...

If NaClNaCl is doped with 10−3mol 6mu%SrCl2,10^{- 3}mol\mspace{6mu}\% SrCl_{2}, then the

concentration of cation vacancies will be.

A

1×10−3mol%1 \times 10^{- 3}mol\%

B

2×10−3mol%2 \times 10^{- 3}mol\%

C

3×10−3mol%3 \times 10^{- 3}mol\%

D

4×10−3mol%4 \times 10^{- 3}mol\%

Answer

1×10−3mol%1 \times 10^{- 3}mol\%

Explanation

Solution

As each Sr2+Sr^{2 +} ion introduces one cation vacancy,

therefore concentration of cation vacancies = mol % of SrCl2SrCl_{2} added.