# Question #df325

##### 1 Answer

#### Answer:

#### Explanation:

You know that the mass of an *individual* lead atom is equal to

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 **each**

Now, your cube has a volume of

#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

#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

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