Question

{"given":"NaCl is doped with $2 \\times 10^{-3}$ mole% $\\mathrm{SrCl_2}$. This means 100 moles of NaCl contain $2 \\times 10^{-3}$ moles of $\\mathrm{SrCl_2}$. Avogadro's number $N_A = 6 \\times 10^{23}$. We need to find the number of cation vacancies per mole of NaCl.","key_observation":"When $\\mathrm{SrCl_2}$ is doped into NaCl, each $\\mathrm{Sr}^{2+}$ ion (charge $+2$) replaces two $\\mathrm{Na}^+$ sites to maintain electrical neutrality. One of these sites is occupied by $\\mathrm{Sr}^{2+}$ itself, and the other site is left vacant. Therefore, each $\\mathrm{Sr}^{2+}$ ion introduced creates exactly one cation vacancy. The total number of cation vacancies equals the total number of $\\mathrm{Sr}^{2+}$ ions introduced per mole of NaCl, multiplied by $N_A$.","option_analysis":[{"label":"(A)","text":"$6.0 \\times 10^{18}$ mol$^{-1}$","verdict":"incorrect","explanation":"This value equals half of the correct answer. It would result if one erroneously assumed each $\\mathrm{Sr}^{2+}$ creates only half a vacancy, or made an arithmetic error by dividing by 2 instead of multiplying."},{"label":"(B)","text":"$1.20 \\times 10^{19}$ mol$^{-1}$","verdict":"correct","explanation":"Moles of $\\mathrm{SrCl_2}$ per mole of NaCl $= \\frac{2 \\times 10^{-3}}{100} = 2 \\times 10^{-5}$ mol. Each $\\mathrm{Sr}^{2+}$ creates one cation vacancy. Number of vacancies $= 2 \\times 10^{-5} \\times 6 \\times 10^{23} = 1.20 \\times 10^{19}$ mol$^{-1}$."},{"label":"(C)","text":"$3.0 \\times 10^{18}$ mol$^{-1}$","verdict":"incorrect","explanation":"This value is approximately one-fourth of the correct answer. It does not correspond to any physically meaningful calculation in this doping problem and likely arises from a combination of arithmetic errors."},{"label":"(D)","text":"$1.20 \\times 10^{21}$ mol$^{-1}$","verdict":"incorrect","explanation":"This value is 100 times larger than the correct answer. It likely arises from incorrectly using $2 \\times 10^{-3}$ moles directly (ignoring the 'mole percent' factor of dividing by 100) and then multiplying by $N_A$."}],"answer":"(B)","formula_steps":[]}