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The distance between Na+ and Cl- ions in solid NaCl of density 43.1 g cm-3 is ________ × 10-10 m. (Nearest Integer)
(Given: NA = 6.02 × 1023 mol-1)
(Given: NA = 6.02 × 1023 mol-1)
Correct answer is '1'. Can you explain this answer?
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The distance between Na+and Cl-ions in solid NaCl of density 43.1 g cm...
The molar mass of NaCl is 58.44 g/mol.
The crystal structure of NaCl is face-centered cubic (FCC), meaning that each Na+ ion is surrounded by 6 Cl- ions and each Cl- ion is surrounded by 6 Na+ ions.
The edge length of the unit cell, a, can be calculated using the formula:
a = (4M/ρN)^1/3
where M is the molar mass of NaCl, ρ is the density of NaCl, and N is Avogadro's number (6.022 x 10^23).
Plugging in the values, we get:
a = (4(58.44)/43.1(6.022 x 10^23))^1/3
a = 5.64 x 10^-8 cm
Since each Na+ ion is surrounded by 6 Cl- ions, the distance between Na and Cl- ions can be calculated as:
distance = a/2
distance = 2.82 x 10^-8 cm
Therefore, the distance between Na and Cl- ions in solid NaCl of density 43.1 g cm^-3 is 2.82 x 10^-8 cm.
The crystal structure of NaCl is face-centered cubic (FCC), meaning that each Na+ ion is surrounded by 6 Cl- ions and each Cl- ion is surrounded by 6 Na+ ions.
The edge length of the unit cell, a, can be calculated using the formula:
a = (4M/ρN)^1/3
where M is the molar mass of NaCl, ρ is the density of NaCl, and N is Avogadro's number (6.022 x 10^23).
Plugging in the values, we get:
a = (4(58.44)/43.1(6.022 x 10^23))^1/3
a = 5.64 x 10^-8 cm
Since each Na+ ion is surrounded by 6 Cl- ions, the distance between Na and Cl- ions can be calculated as:
distance = a/2
distance = 2.82 x 10^-8 cm
Therefore, the distance between Na and Cl- ions in solid NaCl of density 43.1 g cm^-3 is 2.82 x 10^-8 cm.
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Community Answer
The distance between Na+and Cl-ions in solid NaCl of density 43.1 g cm...
Unit cell formula = Na4Cl4
Mass per unit cell =
⇒ a3 = 9.02 × 10-24 cm3
⇒ a = 2.08 × 10-8 cm
⇒ a = 2.08 × 10-10 m
Also, a = 2
⇒
= 1.04 × 10-10 m
∴ The answer is 1.
Mass per unit cell =
⇒ a3 = 9.02 × 10-24 cm3
⇒ a = 2.08 × 10-8 cm
⇒ a = 2.08 × 10-10 m
Also, a = 2
⇒
= 1.04 × 10-10 m∴ The answer is 1.
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The distance between Na+and Cl-ions in solid NaCl of density 43.1 g cm-3is ________ × 10-10m. (Nearest Integer)(Given: NA= 6.02 × 1023mol-1)Correct answer is '1'. Can you explain this answer?
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