An estimation of the radius of a lead atom:
a. You are given a cube of lead that is 1.000 em on each side. The density of lead is
11.35 g/cm3. How many atoms of lead are contained in the sample.
b. Atoms are spherical; therefore, the lead atoms in this sample cannot fill all the
available space. As an approximation, assume that 60% of the space of the cube
is filled with spherical lead atoms. Calculate the volume of one lead atom from
this information. From the calculated volume (V), and the formula V = 47tr3/3, estimate the radius (r) of a lead atom.
c. Assume the lead atoms along each side of the cube touch one another. How many atoms lie along one edge of the cube?
is suppose to be V=4pir^3/3
sry. I get part A. but part B and C are confusing!
a. density = 11.35 g/cc. volume = 1 cm^3. grams = volume x density = ?
Remember there are 6.022E23 atoms in a mole. You have ?g/atomic mass Pb moles Pb atoms.
b. If the volume is 1 cc, then the volume of the atom must be 0.60 if it occupies 60% of the volume. Then volume of 1 Pb atom is
1/# Pb atoms in the 1 cc = xx
I have no idea what 47tr3/3 means to you (nothing to me). You probably meant Volume of a sphere = (4/3)*pi*r^3 and you can use that to calculate the radius.
Post your work if you get stuck.
a. To calculate the number of atoms in the sample, we need to find the mass of the lead cube first. Since we know the dimensions of the cube and its density, we can calculate its mass using the formula:
Density = Mass / Volume
Given that the density of lead is 11.35 g/cm³ and the volume of the cube is (1.000 cm)³, we can rearrange the formula to solve for mass:
Mass = Density * Volume
Mass = 11.35 g/cm³ * (1.000 cm)³
Mass = 11.35 g
We also know that one mole of any element contains Avogadro's number (6.022 x 10²³) of atoms. The molar mass of lead is approximately 207.2 g/mol. To find the number of lead atoms in the sample, we can use the formula:
Number of atoms = (Mass / Molar mass) * Avogadro's number
Number of atoms = (11.35 g / 207.2 g/mol) * (6.022 x 10²³ atoms/mol)
b. To estimate the volume of one lead atom, we can use the given information that the lead atoms fill approximately 60% of the space in the cube. This means that the volume of the cube is 60% occupied by lead atoms, and the remaining 40% is empty space.
Volume of the cube = (1.000 cm)³ = 1.000 cm³
Volume occupied by lead atoms = 60% of 1.000 cm³ = 0.60 * 1.000 cm³
We can now calculate the volume of one lead atom by using the formula:
Volume of one lead atom = (Volume occupied by lead atoms) / (Number of lead atoms)
Note: In this calculation, we assume that the lead atoms are tightly packed and do not account for any intermolecular space.
From the calculated volume (V) and the formula V = (4/3) * pi * r³, we can estimate the radius (r) of a lead atom.
c. If the lead atoms along each side of the cube touch one another, it means that the side of the cube is equal to the diameter of a lead atom. To find the number of atoms along one edge of the cube, we divide the length of the cube by the diameter of a lead atom.
Number of atoms along one edge of the cube = Length of the cube / Diameter of a lead atom
Note: In this calculation, we assume no gaps or overlapping between lead atoms.
a. To find the number of atoms of lead in the sample, we can use the concept of molar mass and Avogadro's number.
First, we need to convert the density of lead from g/cm3 to g/mL since the sample is given in cm (1 mL = 1 cm3).
Density of lead = 11.35 g/cm3 = 11.35 g/mL
Next, we can calculate the mass of the sample:
Mass = Density * Volume
Volume = (1 cm) * (1 cm) * (1 cm) = 1 cm3 = 1 mL
Thus,
Mass = 11.35 g/mL * 1 mL = 11.35 g
Now, we need to calculate the number of moles of lead in the sample using the molar mass of lead.
Molar mass of lead (Pb) = 207.2 g/mol (approximately)
Number of moles = Mass / Molar mass
Number of moles = 11.35 g / 207.2 g/mol = 0.0548 mol (approximately)
Finally, we can use Avogadro's number to find the number of atoms of lead.
Avogadro's number (NA) = 6.022 x 10^23 atoms/mol
Number of atoms = Number of moles * Avogadro's number
Number of atoms = 0.0548 mol * 6.022 x 10^23 atoms/mol = 3.30 x 10^22 atoms (approximately)
Therefore, there are approximately 3.30 x 10^22 atoms of lead in the sample.
b. To estimate the volume and radius of a lead atom, we know that only 60% of the space in the cube is filled with spherical lead atoms.
Volume of the lead atoms = 60% of the volume of the cube
The volume of the cube is given by (side length)3 = (1 cm)3 = 1 cm3.
Volume of lead atoms = 60% of 1 cm3 = 0.6 cm3
Now, we can use the formula for the volume of a sphere to find the radius of a lead atom.
V = (4/3) * π * r3
0.6 cm3 = (4/3) * π * r3
We can rearrange the equation to solve for the radius (r):
r3 = (0.6 cm3) / [(4/3) * π] = 0.45 cm3 / π
r ≈ (0.45 cm3 / π)^(1/3)
Using a calculator, we find that the radius of a lead atom is approximately 0.407 cm.
c. If the lead atoms along each side of the cube touch one another, it means that the distance covered by a single edge is equal to the combined diameter of two lead atoms.
Let's denote the number of lead atoms along one edge of the cube as N.
The distance covered by a single edge of the cube is given by the side length, which is 1 cm.
The diameter of a lead atom is twice the radius, which we found to be approximately 0.407 cm.
Therefore, 1 cm = N * (2 * 0.407 cm)
Simplifying the equation, we can solve for N:
N ≈ 1 cm / (2 * 0.407 cm)
Using a calculator, we find that approximately 1.23 lead atoms lie along one edge of the cube. Since atoms cannot be divided, we can assume there are either 1 or 2 lead atoms along one edge.