The concentration of cation vacancies in NaCl crystal doped with CdCl2 is found to be 6-02- 1016 mol-1- What is the concentration of CdCl2 added to it-

No. of cation vacancies = 6.02 × 10 ^ 16 mol ^ 1 NaCl = 10 ^ 7 × 6.02× 10 ^ 23 mol ^ 1 NaCl ...

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Zero Order Reaction

The concentra...

Question

The concentration of cation vacancies in $N a C l$ crystal doped with $C d C l_{2}$ is found to be $6.02 \times 10^{16} m o l^{- 1}$ . What is the concentration of $C d C l_{2}$ added to it?

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If NaCl is doped with $10^{- 4} m o l % S r C l_{2}$ , the concentration of cation vacancies will be $(N_{A} = 6.02 \times 10^{23} m o l^{- 1})$

10^{- 4} m o l % o f S r C l_{2}

N = 6.02 \times 10^{23} m o l^{- 1}

N a C l

10^{- 4} m o l %

S r C l_{2}

(N_{A} = 6.02 \times 10^{23} m o l^{- 1})

N a C l

10^{- 4}

S r C l_{2}

[N_{A} = 6.02 \times 10^{23} m o 1^{- 1}]

10^{- 4}

S r C l_{2}

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