# Question #4e55d

##### 1 Answer

#### Answer:

#### Explanation:

The first thing to do here is convert your values to *liters* and *grams*, respectively. This will make the calculations *a lot* easier as we go.

So, to convert *gallons* to *liters* and *pounds* to *kilograms*, use the following conversion factors

#"1 gal " ~~ " 3.7854 L"" "# and#" " "1 kg " ~~ " 2.2046 lbs"#

To go from *kilograms* to *Grams*, use the conversion factor

#"1 kg" = 10^3"g"#

So, you will have

#660 color(red)(cancel(color(black)("gal"))) * "3.7854 L"/(1color(red)(cancel(color(black)("gal")))) = "2498.4 L"#

#9.4color(red)(cancel(color(black)("lbs"))) * (1color(red)(cancel(color(black)("kg"))))/(2.2046color(red)(cancel(color(black)("lbs")))) * (10^3"g")/(1color(red)(cancel(color(black)("kg")))) = "4263.8 g"#

Now, the problem wants you to find the concentration of *aluminium cations*, **parts per million**, **ppm**. In order to find the concentration of a solute in ppm, you must essentially figure out how many grams of that solute you get in **solvent**.

#color(blue)(|bar(ul(color(white)(a/a)"ppm" = "grams of solute"/"grams of solvent" xx 10^6color(white)(a/a)|)))#

So, if you have

Now, aluminium sulfate, **per** **percent composition**.

To do that, use the **molar mass** of aluminium and the **molar mass** of aluminium sulfate

#"For Al: " " " " " " "M_M = "26.98 g mol"^(-1)#

#"For Al"_2("SO"_4)_3:" " M_M = "342.15 g mol"^(-1)#

So, the percent composition of aluminium sulfate is

#(color(red)(2) xx 26.98 color(red)(cancel(color(black)("g mol"^(-1)))))/(342.15color(red)(cancel(color(black)("g mol"^(-1))))) xx 100 = "15.77% Al"#

This means that

#4263.8color(red)(cancel(color(black)("g Al"_2("SO"_4)_3))) * overbrace("15.77 g Al"^(3+)/(100color(red)(cancel(color(black)("g Al"_2("SO"_4)_3)))))^(color(purple)("15.77% Al")) = "672.4 g Al"^(3+)#

To get the mass of solvent, which in your case is water, use water's **density**. If no information is provided, you can assume it to be equal to

Remember that

#2498.4color(red)(cancel(color(black)("L"))) * (10^3color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * overbrace("1 g"/(1color(red)(cancel(color(black)("mL")))))^(color(purple)("water's density")) = 2.4984 * 10^6"g"#

So, you know how many grams of aluminium cations you have and how many grams of water you have, which means that you can now find the concentration in ppm

#"ppm" = (672.4 color(red)(cancel(color(black)("g"))))/(2.4984 * color(blue)(cancel(color(black)(10^6)))color(red)(cancel(color(black)("g")))) * color(blue)(cancel(color(black)(10^6))) ="269.13 ppm"#

Rounded to two **sig figs**, the answer will be

#"ppm Al"^(3+) = color(green)(|bar(ul(color(white)(a/a)"270 ppm"color(white)(a/a)|)))#