Question

Please answer 2,3,4

2. How many platinum (Pt) atoms are in a cube that is I mm on each side? What is the average volume occupied by each atom? (Divide the total volume by the number of atoms in the volume.) 3. What is the side of a cube holding exactly 1 mol of gold (Au) atoms? Assume that the equilibrium separation distance of the ions in sodium fluoride (NaF) is the sum of the Na and F ionic radii. What is the Coulombic force of attraction between the two ions? 4.

Answer 1

Volume of cube = (1 mm)^3 = 1 mm^3 density of pt = 21.45 g/cm^3 = s v = 1 mm^3 = 1000 cm^3 mass of pt = p times v = 21.45 g/cm^3 times 1000 cm^3 m = 21.450 g then mole of pt(n) = 21.450 g/195 g/mol = 110 mol then No.of pt atoms.N = n times NA N = 110 mol times (6.022 times 10^23 atom/mol) N = 6.62 times 10^25 atoms Average volume occupied = V/N = 1 mm^3/6.62 times 10^25 atom v_avg = 1.5 times 10^-26 mm^3 Atomic mass of Av = 196 g/mol 1 mol AV = 196 g AV density of AV = 19.32 g/cm^3 Volume = mass/density = 196 g/19.32 g/cm^3 v = 10.15 cm^3 \r\n r_Na^+ = 102 pm, 6F^- = 133 pm then separation.6 = 102 pm + 133 pm = 235 pm 6 = 235 times 10^-12 m since coloumb force of attraction. F = He.e/6 = ke^2/