Zinc Oxide adopts the zinc blende structure and is oberservedto have a density o
ID: 690543 • Letter: Z
Question
Zinc Oxide adopts the zinc blende structure and is oberservedto have a density of 4.86 g/cm3. What is the edge lengthof a unit cell? (I took the molar mass of ZnO and divided it by the density togive an answer in cm3/mol. From there I divided byAvogadro's Number to get cm3/atom. Do Iessentially have a volume for the unit cell? And if Ido, should I use Vcell= (16/3)3 xr3 to find the radius of... And this is where I'm confused. I don't know if this answer would be the radius of a Zn atom or theradius of the O atom.) Zinc Oxide adopts the zinc blende structure and is oberservedto have a density of 4.86 g/cm3. What is the edge lengthof a unit cell? (I took the molar mass of ZnO and divided it by the density togive an answer in cm3/mol. From there I divided byAvogadro's Number to get cm3/atom. Do Iessentially have a volume for the unit cell? And if Ido, should I use Vcell= (16/3)3 xr3 to find the radius of... And this is where I'm confused. I don't know if this answer would be the radius of a Zn atom or theradius of the O atom.) to find the radius of... And this is where I'm confused. I don't know if this answer would be the radius of a Zn atom or theradius of the O atom.)Explanation / Answer
zinc blende has fcc structure. So the no .of atoms per unit cell is 4 Let a be the edge length of te unit cell .Therefore the Volumeof unit cell , V = a^3 Molar mass of ZnO is , M = 65.38 + 16 = 81.38 g Mass of one atom,M = Molar mass / Avagadro number = 81.38 / 6.023 * 10^23 = 13.511 * 10^-23 g / mol Density of the unit cell , d = Mass of the unit cell,M' / Vol.of the unit cell So Volume of unit cell , V = M' / d = 4* M / d = 11.1206 * 10^-23 cm^3 But Volume of unit cell , V = a^3 = 11.1206 *10^-23 cm^3 a = 4.808 * 10^-8 cm = 4.808 * 10^-8 * 10^-2 * 10^2 cm = 480.2 * 10^-10 cm = 480.2 pmRelated Questions
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