Suppose that the atoms in Parts 1, 2 and 3 have an atomic mass of 75.00 g/mol an

Question

Suppose that the atoms in Parts 1, 2 and 3 have an atomic mass of 75.00 g/mol and a radius of 130. pm. Now calculate the unit cell volume in cm^3 and the density (in g/cm^3) of the crystal in each of the three structure types from Parts 1. 2, and 3. Remember that density is the total mass divided by the total volume. Compute the radius of a paladium atom (Pd, atomic number 46) in A (1 A = 10^10 m) from the density of paladium = 22.56 g/cm^3. The unit cell is face-centered cubic. For CsCI, calculate the volume of one unit cell, calculate the mass of one unit cell, and calculate the density of CsCI using the true radii values (from the diameters found in the construction supplement). Note, atoms touch along the body-diagonal but not along the edge in the actual CsCI structure. Magnesium oxide adopts the rock salt (NaCl type) structure, with the distance between the center of the magnesium ion and the center of an pendent oxide ion at 210.6 pm. Calculate the density in g/cm^2 of an ideal MgO crystal from this information and from the data for the model you constructed. Rhenium (_75Re) crystallizes in the help structure. The dimensions of the unit cell are height, 445.6 pm; length of an edge, 276.1 pm. A regular hexagon consists of 6 equilateral triangles; if one edge of the hexagon is length a. then the area of the entire hexagon is (3 Squareroot 3 a^2)/2. Calculate the density of solid rhenium. Refer to the Part 6 model you constructed. lard spheres of radius R are arranged in contact in the form of a simple cubic unit cell. Calculate in terms of R the radius of the largest sphere that can fit in the hole at the center of each face of this unit cell. Analysis Questions: These questions serve in place of an analysts paragraph. Number each answer, and use complete sentences in paragraph style for each question. This can be typed or done in the lab notebook. Make a table that compares simple cubic, face-centered cubic, and body-centered cubic structures in terms of their coordination number, unit cell volume in terms of R, density (see Q.1), and fraction of empty space. State and explain any trends. In general for monoatomic crystals, what two properties of the atom governs the density of the substance? So why are the densities in Q.1 different even though the atoms are the same? Compare and contrast the NaCl unit cell and the face-centered cubic cell. low is the CsCI structure related to the simple cubic cell you constructed in Part 1? 1 low is CsCI different from the body-centered cubic cell in Part 3?

Explanation / Answer

Q2. Palladium (Pd) atom

mass of 4 atoms in fcc unit cell = 106.42 x 4/6.023 x 10^23 = 7.07 x 10^-22 g

Volume of unit cell = 7.07 x 10^-22 g/22.56 g/cm^3 = 3.13 x 10^-23 cm^3

Edge length (d) = cube rt.(3.13 x 10^-23) = 3.15 x 10^-8 cm

Radius = 3.15 x 10^-8/2 x sq.rt.(2)

             = 1.11 x 10^-8 cm = 1.11 Angstrom

Q3. radius of CsCl = 0.161 + 0.181 = 0.348 nm = 3.48 x 10^-8 cm

Volume of unit cell = (4/3)pir^3 = 4 x 3.14 x (3.48 x 10^-8)^3/3

                               = 1.76 x 10^-22 cm^3

Mass of unit cell = 168.36 x 2/6.023 x 10^23 = 5.60 x 10^-22 g

density of CsCl = 5.60 x 10^-22/1.76 x 10^-22 = 3.20 g/cm^3

Q4. NaCl has fcc crystal lattice

mass of unit cell of MgO = 40.30 x 4/6.023 x 10^23 = 2.68 x 10^-22 g

edge length (d) = r x 2(sq.rt(2)) = 2.106 x 10^-8 x 2 x sq.rt.(2) = 5.96 x 10^-8 cm

Volume of unit cell = (5.96 x 10^-8)^2 = 2.116 x 10^-22 cm^3

density of MgO = 2.68 x 10^-22/5.96 x 10^-22 = 1.27 g/cm^3

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