The molar volume of acertain form of solid lead is 18 cm 3 /mol. Assumingcubic c

Question

The molar volume of acertain form of solid lead is 18 cm3/mol. Assumingcubic closest packed structure, determine thefollowing: The radius of a Pbatom. 1.74 pm 17.4 pm 174 pm 1740 pm none of these The unit cell in acertain lattice consists of a cube formed by an anion at eachcorner, an anion in the center, and a cation at the center of eachface. The unit cell contains a net: 5 anions and 6cations 5 anions and 3cations 2 anions and 3cations 3 anions and 4cations 2 anions and 2cations The molar volume of acertain form of solid lead is 18 cm3/mol. Assumingcubic closest packed structure, determine thefollowing: The radius of a Pbatom. 1.74 pm 17.4 pm 174 pm 1740 pm none of these The molar volume of acertain form of solid lead is 18 cm3/mol. Assumingcubic closest packed structure, determine thefollowing: The radius of a Pbatom. 1.74 pm 17.4 pm 174 pm 1740 pm none of these The unit cell in acertain lattice consists of a cube formed by an anion at eachcorner, an anion in the center, and a cation at the center of eachface. The unit cell contains a net: 5 anions and 6cations 5 anions and 3cations 2 anions and 3cations 3 anions and 4cations 2 anions and 2cations Chromium metal crystallizes as abody-centered cubic lattice. If the atomic radius of Cr is1.25 angstroms, what is the density of Cr metal ing/cm3 5.52 7.18 7.81 2.76 3.59

Explanation / Answer

a)                       For CCP , r = 2 / 4 a      Given thatmolar  volume      = 18 cm3/mol      Then volume of oneatom     =  18  cm3 / 6.023 x 10 23                                                 =   2.98 x 10-23 cm3                                                 =    0.298 x 10-24 cm3        This is nothing buta3 . Then a = ( 0.298 x10-24cm3)1/3                                                               = 0.668 x 10-8 cm        This is nothing buta3 . Then a = ( 0.298 x10-24cm3)1/3                                                     r    = 1.414 x 0.668 x 10-8 cm /4                                                       = 0.236 x 10-8 cm                                                       = 23.6 pm b) Anions are in BCC and cations are in FCC.     Total anions = 8 x 1/8 + 1                         =2     Total cations = 6 x 1/2                         =3 2 anions and 3 cations arepresent c)   D = NM/NA a3    For BCC , r = 3/4 a                     a = 4r / 1.732                        = 4 x 1.25 x 10 -8 cm / 1.732                         = 2.88x 10 -8 cm                  N   = 2 atoms ( for BCC )                   M  =51.996 g / mol                              D   = 2 atoms x 51.996 g / 6.023 x 1023 x (2.88 x 10 -8cm )3                       = 0.72 x 10 g / cm3                       = 7.2 g /cm3                       = 7.2 g /cm3

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