Question

1. if the density of lead (pb) is 11.34 g/cm³, how many lead atoms are in a rectangular prism of lead that is 2.00 cm wide, 1.00 m long, and 2.00 mm height? given = volume of a rectangular prism = width × length × height

a) 1.03×10²² atoms
b) 1.32×10²³ atoms
c) 1.32×10²⁴ atoms
d) 1.16×10²³ atoms
e) 6.60×10²³ atoms

Explanation:

Step1: Convert dimensions to cm

The width $w = 2.00\ cm$, the length $l=1.00\ m = 100\ cm$, and the height $h = 2.00\ mm=0.200\ cm$.

Step2: Calculate the volume of the rectangular - prism

The volume formula for a rectangular prism is $V = w\times l\times h$. So $V=(2.00\ cm)\times(100\ cm)\times(0.200\ cm)=40.0\ cm^{3}$.

Step3: Calculate the mass of lead

Using the density formula $
ho=\frac{m}{V}$, we can find the mass $m$. Given $
ho = 11.34\ g/cm^{3}$ and $V = 40.0\ cm^{3}$, then $m=
ho V=(11.34\ g/cm^{3})\times(40.0\ cm^{3}) = 453.6\ g$.

Step4: Calculate the number of moles of lead

The molar mass of lead ($Pb$) is $M = 207.2\ g/mol$. The number of moles $n=\frac{m}{M}=\frac{453.6\ g}{207.2\ g/mol}=2.19\ mol$.

Step5: Calculate the number of lead atoms

Using Avogadro's number $N_{A}=6.022\times 10^{23}\ atoms/mol$, the number of atoms $N=nN_{A}=(2.19\ mol)\times(6.022\times 10^{23}\ atoms/mol)=1.32\times 10^{24}\ atoms$.

Answer:

C. $1.32\times 10^{24}$ atoms