Question

The distance between Na⁺ and Cl⁻ ions in solid NaCl of density 43.1 g cm⁻³ is ______ × 10 ⁻¹⁰ m (Nearest integer).

(Given: N_A = 6.02 × 10²³ mol⁻¹)

Answer 1

The distance between Na⁺ and Cl⁻ ions in solid NaCl of density 43.1 g cm⁻³ is 1 × 10 ⁻¹⁰ m.

Explanation:

Given: Density of NaCl = 43.1 g cm⁻³

N_A = 6.02 × 10²³ mol⁻¹

Molar mass of NaCl = 58.5 g mol⁻¹

Formula: ρ = `(Z xx M )/(N_A xx a^3)`

ρ = `(4 xx 58.5)/(N_A xx a^3) `

⇒ `(4 xx 58.5)/(6.02 × 10^23 xx a^3) = 43.1`

⇒ `(234)/(43.1 xx 6.02 × 10^23) = a^3`

⇒ `(234)/(259.462 × 10^23) = a^3`

⇒ a³ = 0.902 × 10⁻²³ cm³

⇒ a³ = 9.02 × 10⁻²⁴ cm³

⇒ a = `sqrt(9.02 × 10^(−24)  cm^3)`

⇒ a = 2.08 × 10⁻⁸ cm

⇒ a = 2.08 × 10⁻¹⁰ m

The distance between ions is also,

a = 2 (r_Na + r_Cl)

⇒ 2(r_Na + r_Cl) = 2.08 × 10⁻¹⁰ m

⇒ `r_(Na) + r_(Cl) = (2.08 xx 10^-10)/2`

⇒ r_Na + r_Cl = 1.04 × 10⁻¹⁰ m

= 1.04 × 10⁻¹⁰ ≈ 1  (∵ Nearest integer is 1.)