# Question df325

Nov 22, 2015

$6.6 \cdot {10}^{22} \text{atoms}$

#### Explanation:

You know that the mass of an individual lead atom is equal to $3.4 \cdot {10}^{- 22}$ grams.

A substance's density tells you what its mass per unit of volume is. In this case, you know that lead has a density of ${\text{11.3 g/cm}}^{3}$, which means that each ${\text{cm}}^{3}$ of lead will have a mass of $\text{11.3 g}$.

Now, your cube has a volume of ${\text{2.00 cm}}^{3}$, which means that its mass will be equal to

2.00color(red)(cancel(color(black)("cm"^3))) * overbrace("11.3 g"/(1color(red)(cancel(color(black)("cm"^3)))))^(color(blue)("density")) = "22.6 g"

Now all you have to do is look at the mass of one lead atom and figure out exactly how many lead atoms would be needed to make the total mass of the cube equal to $\text{22.6 g}$.

22.6color(red)(cancel(color(black)("g"))) * overbrace("1 lead atom"/(3.4 * 10^(-22)color(red)(cancel(color(black)("g")))))^(color(blue)("the mass per atom")) = 6.647 * 10^(22)"lead atoms"

Rounded to two sig figs, the number of sig figs you have for the mass of an individual lead atom, the answer will be

"no. of lead atoms" = color(green)(6.6 * 10^(22)"atoms")

You can double-check this result by using lead's molar mass, which tells you what the exact mass of one mole of lead is.

22.6color(red)(cancel(color(black)("g"))) * "1 mole Pb"/(207.2color(red)(cancel(color(black)("g")))) = "0.1091 moles Pb"

Now use Avogadro's number to determine how many atoms of lead you have in that many moles

0.1091color(red)(cancel(color(black)("moles"))) * (6.022 * 10^(23)"atoms")/(1color(red)(cancel(color(black)("mole")))) = 6.57 * 10^(22)"atoms"#

To two sig figs, you will once again get

$\text{no. of lead atoms" = color(green)(6.6 * 10^(22)"atoms}$