Question #10a69

Your starting point here will be the chemical formula of cisplatin

`"Pt"("NH"_3)_2"Cl"_2`

Notice that every cisplatin coordination complex contains `2` atoms of chlorine. This means that in order to have `524` atoms of chlorine, you need

`524 color(red)(cancel(color(black)("atoms Cl"))) * "1 cisplatin complex"/(2color(red)(cancel(color(black)("atoms Cl")))) = "262 cisplatin complexes"`

Now, cisplatin has a molar mass of `"330.01 g mol"^(-1)`, which means that every mole of cisplatin has a mass of `"300.01 g"`.

Since mole of cisplatin contains `6.022 * 10^(23)` cisplatin complexes, you can say that `262` complexes will be equivalent to

`262 color(red)(cancel(color(black)("cisplatin complexes"))) * overbrace("1 mole cisplatin"/(6.022 * 10^(23)color(red)(cancel(color(black)("cisplatin complexes")))))^(color(blue)("Avogadro's constant"))`

`= 4.351 * 10^(-22)color(white)(.)"moles cisplatin"`

You can thus say that the mass of cisplatin that contains `524` atoms of chlorine is equal to

`4.351 * 10^(-22) color(red)(cancel(color(black)("moles cisplatin"))) * overbrace("300.01 g"/(1color(red)(cancel(color(black)("mole cisplatin")))))^(color(blue)("the molar mass of cisplatin"))`

`= color(darkgreen)(ul(color(black)(1.31 * 10^(-19)color(white)(.)"g")))`

The answer is rounded to three sig figs.