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Sodium chloride crystallizes in face-cen...

Sodium chloride crystallizes in face-centred cubic (f.c.c.) structure. Its density is 2.165 g `cm^(-3)`. If the distance between `Na^+` and its nearest `Cl^(-)` ions is 281 pm, find out the Avogadro's number (Na = 23 g `"mol"^(-), Cl = 35.5 g "mol"^(-)`).

Text Solution

Verified by Experts

For a cubic unit cell,
`d=(ZM)/(a^3 10^(-30)N_A) g cm^(-3)` or `N_A=(ZM)/(10^(-30)d a^3)`
When .a. is in pm.
Where M=Molar mass of NaCl = 23 + 35.5 = 58.5 g `mol^(-)`
`N_A`=Avogadro.s number = ?
d= density = 2.165 g `cm^(-3)`
Z= number of atoms per unit cell = 4 (f.c.c.)
a=edge length of the unit cell in pm.
For NaCl type crystal, edge length (a) = `2 xx "interionic distance"` = 2 (281) = 562 pm
`:. N_A=(4 xx 58.5 g mol^(-))/(10^(-30) xx 2.165 g cm^(-3) xx (562)^3 cm^3)`
`=(234.0 g mol^(-1))/(8 xx 22188041 xx 10^(-30) xx 2.165 g) = 6.09 xx 10^(23) mol^(-)`
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Knowledge Check

  • The distance between Na^+ and Cl^- ions in NaCl with a density 3.165 g cm^(-3) is

    A
    497 pm
    B
    248.5 pm
    C
    234 pm
    D
    538.5 pm

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