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If NaCl is doped with \small 10^{-3} mol% GaCl3 , what is the concentration of the cation vacancies per mole of NaCl?

Option: 1

1.205 \times 10^{15}

Option: 2

1.2 \times 10^{19}

Option: 3

12.05 \times 10^{19}

Option: 4

15.43 \times 10^{17}

Answers (1)


We have,
100 moles of NaCl are doped with 10^{-3} moles of  GaCl_3 .
Thus, 1 mole of NaCl is doped with GaCl_3=10^{-5} moles

Now, as we know, that as one Ga3+ ion is introduced, three Na+ needs to be removed to maintain the electrical neutrality. So as one vacancy is filled by Ga3+ ion , two cation vacancies are formed.

Therefore, concentration of cation vacancy =2\times 10^{-5} moles/mole of NaCl
Thus, the number of Cationic vacancy =2 \times 10^{-5} \times 6.023 \times 10^{23} per mole of NaCl

Hence, concentration of cation vacancies =1.2 \times 10^{19} per mole of NaCl

Therefore, Option(2) is correct

Posted by

Kuldeep Maurya

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