Verified Solution • 2025–26 Session moderate

QAIPMT-2007-Ch-2317: NEET 2007 Chemistry - Class 12 AIPMT-2007-Ch-2317 Solution: If NaCl is doped with \(10^{-4}\) mol% of \(SrCl_2\), the…

If NaCl is doped with 10−410^{-4} mol% of SrCl2SrCl_2, the concentration of cation vacancies will be (NA=6.02×1023 mol−1N_A=6.02\times10^{23}\ \text{mol}^{-1})

Multiple Choice Options

  • A 6.02×1016 mol−16.02\times10^{16}\ \text{mol}^{-1}
  • B 6.02×1017 mol−16.02\times10^{17}\ \text{mol}^{-1}
  • C 6.02×1014 mol−16.02\times10^{14}\ \text{mol}^{-1}
  • D 6.02×1015 mol−16.02\times10^{15}\ \text{mol}^{-1}

Step-by-Step Verified Solution

As each Sr 2+ ion introduces one cation vacancy, therefore concentration of cation vacancies = mole % of SrCl 2 added. _ Concentration of cation vacancies = 10 �4 mole % = 10 _ 4 100 _ 6.023 _ 10 23 = 6.023 _ 10 23 _ 10 -6 = 6.023 _ 10 17

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