Question #10a69

1 Answer
Mar 4, 2017

Answer:

#1.31 * 10^(-19)"g"#

Explanation:

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.