Question

A cube made of platinum (Pt) has an edge length of 1.0 cm. (a) Calculate the number of Pt atoms in the cube. (b) Atoms are spherical in shape. Therefore, the Pt atoms in the cube cannot fill all the available space. If only 74 percent of the space inside the cube is taken up by Pt atoms, calculate the radius in picometers of a Pt atom. The density Pt is 21.45 g/cm3, and the mass of a single Pt atom is 3.240 x 10^-22 g. (The volume of a sphere of radius r is 4/5Ïr^3).

Answer 1

Answer:

a)The total number of platinum atoms are $6.620\times 10^{22} atoms$.

b)The radius of a Pt atom is 138.7 pm.

Explanation:

a) Let the total number of platinum atoms be x

And mass of x atoms of platinum be m

Edge length of platinum cube ,s= 1.0 cm

Volume of the cube = $1.0 cm^3$

Density of the platinum cube = $d=21.45 g/cm^3$

Mass = Density × Volume

$(m)=21.45 g/cm^3\times 1.0 cm^3=21.45 g$

Mass of an atom of platinum = $3.240\times 10^{-22} g$

$x\times 3.240\times 10^{-22} g=21.45 g$

$x=6.620\times 10^{22} atoms$

b) If only 74 percent of the space inside the cube is taken up by Pt atoms.

Volume occupied by platinum atoms = 74% of $1.0 cm^3$

$V'= 0.74 cm^3$

Volume occupied by $6.620\times 10^{22}$ atoms is$0.74 cm^3$.

$6.620\times 10^{22} \times \frac{4}{3}\pi r^3=V'$

$6.620\times 10^{22} \times \frac{4}{3}\times 3.14\times r^3=0.74 cm^3$

$r = 1.387\times 10^{-8} m = 138.7 pm$

The radius of a Pt atom is 138.7 pm.