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What is the distance between Na^(+) and Cl^(-) ions in NaCl crystal if its density is 2.165g cm^(-3) ? NaCl crystallizes in fcc lattice. |
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Answer» Solution :Let the edge length of unit CELL = a cm For fcc structure, number of MOLECULES per unit cell, Z=4 Gram formula mass of NaCl, M = 23 +35.5=58.5g `mol^(-1)` Mass of unit cell = `(Z xx" Gram formula mass")/("Avogadro.s number")` = `(4 X (58.5 g mol^(-1)))/((6.022 xx 10^(23) mol^(-1)) = 3.886 xx 10^(-22)` g Density of unit cell = `("Mass")/("Volume")` or `2.165 g cm^(-3) = (3.886 xx 10^(-22)g)/(a^3)` `:. a^3 =((3.886 xx 10^(-22)g))/((2.165 g cm^(-3)) = 1.795 xx 10^(-22) cm^3` `= 179.5 xx 10^(-24) cm^3` Edge length, `a=(179.5)^(1//3) xx 10^(-8) cm^3` = `5.64 xx 10^(-8)` cm=564 pm Edge length of `Na^(+)Cl^(-)` unit cell = `2 (""^(r)N_a^(+) + ""^(r)Cl_a^(-))`=564 pm or `(""^(r)Na_a^(+) + ""^(r)Cl_a^(-)) =(564)/(2) = 282` pm Distance between `Na^+` and `Cl^(-)` ions = 282 pm |
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