Question

8 grams of barium carbonate (BaCO3) is needed to kill a rat. How many molecules is this?

Answer 1

`~~2.44*10^22` molecules of `BaCO_3`

Explanation:

We first need to convert the amount of barium carbonate into moles.

Barium carbonate `BaCO_3` has a molar mass of `197.34 \ "g/mol"`.

Here, there are `8` grams of barium carbonate, so there are

#(8cancel"g")/(197.34cancel"g""/mol")=0.040539171~~4.05*10^-2 \ "mol of" \ BaCO_3#

We know that, in one mole of molecules, there are going to be approximately `6.02*10^23` molecules.

So here, there exists `4.05*10^-2*6.02*10^23~~2.44*10^22` molecules of `BaCO_3`.