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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"^(-)). |
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Answer» Solution :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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