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an unknown element crystallizes in a face-centered cubic unit cell. with an edge length of 392.4 pm. The solid has a density of 21.09 g/cm^3. what is atomic weight of solid?

so what i did was :

FC cubic uit --> 4 atoms
edge length--> l = 2*root2*r = 392.4 pm = 392.4 * 10^-8 cm.
d= 21.09 g/cm^3
d= m/l^3

m= l^3 * d

m= 8.2757*10^-5 g

1amu = 1.66054 * 10^-24 g

atomic weight = 4.98 * 10^19 amu.

and that is not the correct answer at all. where am i going wrong? please help :/

the options are :

a) 6.9 amu
b) 241.7 amu
c)74.4 amu
d) 191.8 amu

  • chemistry - ,

    ryeng i suppose?
    this is a very easy question
    392.4*10-10= x cm
    xcm^3= ycm^3

    ycm^3* 21.09= mass in grams
    since face centered unit cell has 4 atoms
    mass/4, then use the product of that to divide by 1.66*10^-24 = amu
    your result should be 191.84 amu

  • chemistry - ,

    One place you erred is in the volume. You have a given to you (the edge length) so volume = a^3.

    I agree with Tom's answer although I don't follow all of the work. My answer is 191.77 which rounds to 191.8 and that is one of the answers.

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