Germanium has a face-centered cubic unit cell. The density of germanium is 5.32 g/cm3. Calculate a value for the atomic radius of germanium.

1.59x10⁻¹⁰m

Explanation:

To solve this question we must know that the length of the cubic cell, X, is equal to:

X = √8 * R

Where R is the atomic radius of germanium

And that in 1 unit cell there are 4 atoms of germanium.

To solve this question we must find the mass in 1 unit cell, with this mass we can find the volume of the cube and the length. With the length we can know the atomic radius:

Mass in 1 unit cell -Molar mass Ge = 72.64g/mol:

4 atoms Ge * (1mol / 6.022x10²³ atoms) = 6.64x10⁻²⁴ moles Ge

6.64x10⁻²⁴ moles Ge * (72.64g / mol) = 4.825x10⁻²²g Ge

Volume unit cell:

4.825x10⁻²²g Ge * (1cm³ / 5.32g) = 9.07x10⁻²³cm³

Length unit cell:

∛9.07x10⁻²³cm³ = 4.49x10⁻⁸cm * (1m / 100cm) = 4.49x10⁻¹⁰m

Atomic radius Ge:

4.49x10⁻¹⁰m / √8 =

the ratio of the conjugate base to the acid required to attain the desired ph.

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