Given: NaCl is doped with 10^-2 mole % of SrCl₂
(i.e.) 100 moles of NaCl doped with 10^-2 moles of SrCl₂
\(\therefore \) 1 mole of NaCl is doped with 10^-4 moles of SrCl₂
1 Sr²⁺  ion creates 1 cation vacancy
The number of cation vacancies created by 10^-4 mole SrCl₂ = 10^-4 \(\times\) 6.023 \(\times\) 10²³
= 6.023 \(\times\) 10¹⁹ vacancies

\(\operatorname{Density}(\rho)=\frac{\mathrm{nM}}{\mathrm{a}^{3} \mathrm{~N}_{\mathrm{A}}} \)
\(\mathrm{n}=2, \mathrm{~N}_{\mathrm{A}}=6.023 \times 10^{23} ; \mathrm{a}=288 \mathrm{pm}=2.88 \times 10^{-8} \mathrm{~cm}, \rho=7.2 \mathrm{~g} \mathrm{~cm}^{-3} \)
\(\therefore M=\frac{\rho \times a^{3} \times N_{A}}{n} \)
\(=\frac{7.2 \times\left(2.88 \times 10^{-8}\right)^{3} \times 6.023 \times 10^{23}}{2} \)
\(=517.95 \times 10^{-1} \)
\(=51.795 \mathrm{~g} \mathrm{~mol}^{-1} \)
Number of moles (n) \(=\frac{\text { Mass }}{\text { Molar mass }}=\frac{208}{51.795}\)=4.02 moles
No. of atoms = No. of moles \(\times\) Avogadro number
=n \(\times\) N_A
\(=4.01 \times 6.023 \times 10^{23} \)
\(=24.15 \times 10^{23} \text { atoms }\)

For bcc structure \((r)=\frac{\sqrt{3}}{4} a\)
a = 4.3 \(\times\) 10^-8 cm, r = ?
\(=\frac{1.732 \times 4.3 \times 10^{-8}}{4}\)
\(r=1.86 \times 10^{-8} \mathrm{~cm}\)

\(\operatorname{Density}(\rho)=\frac{\mathrm{nM}}{\mathrm{a}^{3} \mathrm{~N}_{\mathrm{A}}} \)
\(\rho=10 \mathrm{~g} \mathrm{~cm}^{-3} ; \mathrm{a}=100 \mathrm{pm}=1 \times 10^{-8} \mathrm{~cm} ; \mathrm{N}_{\mathrm{A}}=6.023 \times 10^{23} ; \mathrm{n}=4 ; \mathrm{M}=? \)
\(M=\frac{\rho \mathrm{a}^{3} \mathrm{N_{A}}}{n} \)
\(=\frac{10 \times\left(1 \times 10^{-8}\right)^{3} \times 6.023 \times 10^{23}}{4}  \)
\(=\frac{6.023}{4} \)
= 1.505 g /mol
No. of moles \(=\frac{\text { Mass }}{\text { Molar mass }}=\frac{1}{1.505}\)
= 0.664 moles
Hence number of atoms = 0.664 x 6.023 x 10²³ = 3.99 x 10²³ atoms

\(\text { Density }(\rho)=\frac{\mathrm{nM}}{\mathrm{a}^{3} \mathrm{~N}_{\mathrm{A}}}  \)
\(\mathrm{n}=4, \mathrm{M}=\text { Molar mass of } \mathrm{KF}=58.1 \mathrm{~g} / \mathrm{mol}  \)
\(\rho=2.48 \mathrm{~g} \mathrm{~cm}^{-3}  \)
\(\mathrm{~N}_{\mathrm{A}}=6.023 \times 10^{23} \)
\(a^{3} =\frac{n M}{\rho N_{A}}=\frac{4 \times 58.1}{2.48 \times 6.023 \times 10^{23}}  \)
\(a^{3} =15.55 \times 10^{-23}  \)
\(a^{3} =0.1555 \times 10^{-21}  \)
\(a =\sqrt[3]{0.1555 \times 10^{-21}}  \)
\(a =0.5375 \times 10^{-7} \mathrm{~cm}=5.375 \times 10^{-8} \mathrm{~cm}=537.5 \mathrm{pm}  \)
\(d =\frac{a}{\sqrt{2}}(\text { for fcc })  [\therefore r = \frac{a\sqrt{2}}{4}]\)
\(=\frac{537.5}{1.414}=380.13 \mathrm{pm}\)
\(\therefore\) The distance between K⁺ and F^- in KF = 380.13 pm