Homework Solution: Chromium forms a hexagonal close packed crystal structure. Determin…

    Chromium forms a hexagonal close packed crystal structure. Determine the theoretical density of chromium 2. nA n number of atoms per unit cell (atoms): Ve volume of the unit cell (units) Na= avogadros number (atoms/mol); A = atomic weight of the atom: 51.996 g/mol
    Chromium forms a hexagonal close packed crystal structure. Determine the theoretical density of chromium rho = nA/V_cN_a n = number of atoms per unit cell (atoms): V_c = volume of the unit cell (units^3) N_A= avogadro's number (atoms/mol): A = atomic weight of the atom = 51.996 g/mol

    Expert Answer

    Answer:

    Chromium forms a hexagonal cease packed crystal edifice. Determine the speculative dullness of chromium 2. nA n estimate of molecules per item cell (atoms): Ve book of the item cell (units) Na= avogadros estimate (atoms/mol); A = ultimate moment of the molecule: 51.996 g/mol

    Chromium forms a hexagonal cease packed crystal edifice. Determine the speculative dullness of chromium rho = nA/V_cN_a n = estimate of molecules per item cell (atoms): V_c = book of the item cell (units^3) N_A= avogadro’s estimate (atoms/mol): A = ultimate moment of the molecule = 51.996 g/mol

    Expert Tally

    Answer:

    For the HCP packing edifice, we enjoy n = 6

    Book of item cell, Vc = 33.94*r3

    Here, r = radius of Cr molecule = 200 pm = 2*10-10 m

    So,

    Vc = 33.94*(2*10-10)3 = 271.52*10-30 m3.

    Na = 6.02*1023 atoms/mol

    Putting these values we get:

    Density, d = (6*51.996)/(271.52*10-30*6.02*1023) = 1.908*106 g/m3