Question

Argon crystallizes in the face-centered cubic arrangement at 40k. given that the atomic radius of argon is 191 pm, calculate the density of solid argon.

Answer 1

To calculate the density of solid argon at 40 K, we first determine the edge length of the unit cell using the atomic radius, then calculate the volume of the unit cell. We then find the mass of argon in the unit cell using its atomic weight and Avogadro's number and divide the mass by the volume to find the density.

Step-by-step explanation:

To calculate the density of solid argon, we can follow these steps:

1. First, we need to know the edge length a of the face-centered cubic unit cell. Since argon has a face-centered cubic arrangement and the atomic radius is given as 191 pm, we can use the equation for the face-centered cubic structure a = 2√2×r, where r is the atomic radius.
2. Next, we calculate the volume of the unit cell by cubing the edge length: V = a³.
3. Since there are four argon atoms per face-centered cubic unit cell, we multiply the number of atoms by the atomic weight of argon (39.948 g/mol) to get the mass of argon contained within one unit cell.
4. To find the density (ρ), we divide the mass of the unit cell by its volume and then convert the units to the desired kg/m³.
5. Remember to use Avogadro's number (6.022×10²³ mol¹) when converting from grams per mole to grams per unit cell.

By using the appropriate equations and constants, we can find the value for the density of solid argon at 40 K.

Answer 2

Molecular weight of 1 mole of Argon = 39.948 g
for face-centered cube = x = (√8)r and here r = 191 pm
so, x = √8 x 191 = 540 pm = 540 x 10^-10 cm
density = (39.948g / mol) x (unit cell / 540x10^-10 cm)^3 x (mol / 6.022x10^23 atoms) x (4 atoms / unit cell) = 1.69g/cm^3
so density of solid argon is 1.69g/cm^3