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The well known mineral fluorite is chemically calcium fluoride. It is known that in one unit cell of this mineral there are 4 Ca^(2+) ions and 8F^(-) ions and that Ca^(2+) ions are arranged in a fcc-lattice. The F^(-) ions fill all the tetrahedral holes in the face centred cubic lattice of Ca^(2+) ions. The edge of the unit cell is 5.46 xx 10^(-8) cm in length. The density of the solid is 3.18 g cm^(-3). Use this information to calculate Avogadro's number. (Molar mass of CaF_(2) = 78.08 g mol^(-1)) |
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Answer» Solution :`d = (Z xx M)/(a^(3) xx N_(A))` For fcc LATTICE Z = 4 3.18 g `CM^(-3) = (4 xx 78.08 g mol^(-1))/((5.46 xx 10^(-8)cm)^(3) xx N_(A))` `N_(A) = (4 xx 78.08 g mol^(-1))/((5.46 xx 10^(-8)cm)^(3) xx 3.18 g cm^(-3))` `N_(A) = 6.033 xx 10^(23) mol^(-1)` |
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