If NaCl is doped with 10-4 mol - SrCl 2- the concentration of cation vacancies will be N A -6-02 - 1023 mol -1A- 6-02 - 1017 mol -1B- 6-02 - 1016 mol -1C- 6-02 - 1014 mol -1D- 6-02 - 1015 mol -1

The correct option is A 6.02 × 10^17mol^ 1 Number of moles of cationic vacancies 10^ 4/10^2 = 10^ 6 mole ⇒ Number of cationic vacancies = 10^ 6× 6.02 × 1023 = ...

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Standard XII

Chemistry

Voids

If NaCl is do...

Question

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

$6.02 \times 10^{14} m o l^{- 1}$

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$6.02 \times 10^{15} m o l^{- 1}$

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$6.02 \times 10^{16} m o l^{- 1}$

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$6.02 \times 10^{17} m o l^{- 1}$

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Solution

The correct option is D
$6.02 \times 10^{17} m o l^{- 1}$

Number of moles of cationic vacancies

$\frac{10^{- 4}}{10^{2}} = 10^{- 6} m o l e$

$\Rightarrow$ Number of cationic vacancies $= 10^{- 6} \times 6.02 \times 1023 = 6.02 \times 10^{17} m o l^{- 1}$

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Similar questions

Q. If NaCl is doped with

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

, the concentration of cation vacancies will be (

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

)

Q. If NaCl is doped with

10^{- 4}

mole % of

S r C l_{2}

, the concentration of cation vacancies will be :

Q. If

N a C l

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})

Q. If

N a C l

is doped with

10^{- 4}

mole % of

S r C l_{2}

, the concentration of cation vacancies will be:

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

Q. 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 mol% SrCl2, the concentration of cation vacancies will be (NA=6.02×1023mol−1)

The correct option is D 6.02×1017mol−1 Number of moles of cationic vacancies 10−4102=10−6mole ⇒ Number of cationic vacancies =10−6×6.02×1023=6.02×1017 mol−1

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

The correct option is D
$6.02 \times 10^{17} m o l^{- 1}$

Number of moles of cationic vacancies

$\frac{10^{- 4}}{10^{2}} = 10^{- 6} m o l e$

$\Rightarrow$ Number of cationic vacancies $= 10^{- 6} \times 6.02 \times 1023 = 6.02 \times 10^{17} m o l^{- 1}$