The distance between and ions in solid NaCl of density is ____ ×
. (Nearest Integer)
(Given : )
. (Nearest Integer)
(Given : )
Answer: 1
Solution
NaCl has an FCC structure with Z = 4 formula units per unit cell.
Mass per unit cell = (Z × M.M.)/Nₐ = (4 × 58.5)/(6.02 × 10²³) g.
Density d = m/a³, so 43.1 = (4 × 58.5)/(Nₐ × a³).
Solving: a³ = 9.02 × 10⁻²⁴ cm³, giving a = 2.08 × 10⁻⁸ cm = 2.08 × 10⁻¹⁰ m.
In the FCC structure, the distance between nearest Na⁺ and Cl⁻ ions is a/2 = (r_Na⁺ + r_Cl⁻) = 1.04 × 10⁻¹⁰ m.
Therefore, x = 1.
Mass per unit cell = (Z × M.M.)/Nₐ = (4 × 58.5)/(6.02 × 10²³) g.
Density d = m/a³, so 43.1 = (4 × 58.5)/(Nₐ × a³).
Solving: a³ = 9.02 × 10⁻²⁴ cm³, giving a = 2.08 × 10⁻⁸ cm = 2.08 × 10⁻¹⁰ m.
In the FCC structure, the distance between nearest Na⁺ and Cl⁻ ions is a/2 = (r_Na⁺ + r_Cl⁻) = 1.04 × 10⁻¹⁰ m.
Therefore, x = 1.
JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Integer answer