# Question 10a69

Mar 4, 2017

$1.31 \cdot {10}^{- 19} \text{g}$

#### Explanation:

Your starting point here will be the chemical formula of cisplatin

${\text{Pt"("NH"_3)_2"Cl}}_{2}$

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 ${\text{330.01 g mol}}^{- 1}$, which means that every mole of cisplatin has a mass of $\text{300.01 g}$.

Since mole of cisplatin contains $6.022 \cdot {10}^{23}$ 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"))

$= 4.351 \cdot {10}^{- 22} \textcolor{w h i t e}{.} \text{moles cisplatin}$

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"))#

$= \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{1.31 \cdot {10}^{- 19} \textcolor{w h i t e}{.} \text{g}}}}$

The answer is rounded to three sig figs.