Question

Calculate the number of atoms contained in a cylinder (1 m radiusand1 m deep)of (a) magnesium (b) lead.

Answer 1

Answer:

The question is incomplete,below is the complete question

"Calculate the number of atoms contained in a cylinder (1μm radius and 1μm deep)of (a) magnesium (b) lead."

Answer:

a. 1.35*10^{11} atoms

b. 1.03*10^{11} atoms

Explanation:

First, we determine the volume of the magnesium in the cylinder container

using the volume of a cylinder

$V=\pi r^{2}h\\ r=10^{-6}m\\ h=10^{-6}m\\V=\pi *10^{-6*2}*10^{-6}\\V=\pi *10^{-18}\\V=3.14*10^{-18}m^{3}$

a. Next we determine the mass of the magnesium ,

using the density=mass/volume

since density of a magnesium

$the density of magnesium =1.738*10^{3}kg/m^{3} \\mass=density * volume \\mass=1.738*10^{3}*3.14*10^{-18}\\mass=5.46*10^{-15}kg\\ \\mass=5.46*10^{-12}g$

Finally to calculate the number of atoms,

we determine the number of moles

mole=mass/molarmass

$mole=5.46*10^{-12}/ 24.305\\mole=0.225*10^{-12}mol$

Hence the number of atoms is

number of atoms=mole*Avogadro's constant

$number of atoms = 0.225*10^{-12}*6.02*10^{23}\\number of atoms =1.35*10^{11} atoms$

b. for he lead, we determine the mass of the lead  ,

using the density=mass/volume

since density of a magnesium

$the density of lead =11.34*10^{3}kg/m^{3} \\mass=density * volume \\mass=11.34*10^{3}*3.14*10^{-18}\\mass=35.60*10^{-15}kg\\ \\mass=35.60*10^{-12}g$

Finally to calculate the number of atoms,

we determine the number of moles

mole=mass/molarmass

$mole=35.60*10^{-12}/ 207.2\\mole=0.1718*10^{-12}mol$

Hence the number of atoms is

number of atoms=mole*Avogadro's constant

$number of atoms = 0.1718*10^{-12}*6.02*10^{23}\\number of atoms =1.03*10^{11} atoms$