Rubidium crystalizes in the body centred cubic structure with an edge length of 560.5 pm..
a. calculate the radius (in pm) to 4 sig figures.
b. calculate the density to 4 sig.fgures.
thnks guys..
a = edge length = 560.5 pm.
4*radius = a*31/2
Convert a to cm here, I think it is easier here than later.
Then volume = a^3 in cc.
density = mass/volume so we need to calculate mass.
mass of unit cell = #atoms per unit cell*molar mass Rb/6.02E23 = ?
can u please explain to me where did u get the 3 to find radius?..
thank you..
and also how to get the # of atoms per unit cell...
a. To calculate the radius of a body-centered cubic (BCC) structure, we need to use the formula:
Radius = (√3 / 4) * edge length
Given that the edge length is 560.5 pm, we substitute it into the formula:
Radius = (√3 / 4) * 560.5 pm
Radius ≈ 485.34 pm
Therefore, the radius of the Rubidium crystal is approximately 485.34 pm.
b. To calculate the density of a substance, we need to divide its mass by its volume. For the body-centered cubic structure, the volume is given by:
Volume = (edge length)^3
Given that the edge length is 560.5 pm, we substitute it into the formula:
Volume = (560.5 pm)^3
Volume ≈ 1.129 × 10^8 pm^3
Now, we need to determine the mass of Rubidium. The molar mass of Rubidium (Rb) is approximately 85.468 g/mol.
To convert grams to pm^3, we need to use Avogadro's number (6.022 × 10^23) and the molar mass of Rubidium:
Mass = (85.468 g/mol) / (6.022 × 10^23) * 1.66053906660 × 10^-24 g/pm^3
Mass ≈ 2.324 × 10^-22 g/pm^3
Now, we can calculate the density:
Density = Mass / Volume
Density ≈ (2.324 × 10^-22 g/pm^3) / (1.129 × 10^8 pm^3)
Density ≈ 2.056 × 10^-30 g/pm^3
In scientific notation, the density of Rubidium is approximately 2.056 × 10^-30 g/pm^3, rounded to 4 significant figures.