# A sample of seawater contains 6.277 g of sodium chloride per liter of solution. How many mg of sodium chloride would be contained in 15.0 mL of this solution?

##### 1 Answer

#### Answer:

#### Explanation:

You can actually use multiple approaches to solve this problem, but they revolve around a couple of unit conversions.

One way to approach this is to use the given concentration to figure out how many *milligrams* of sodium chloride you get **per milliliter** of solution.

This will allow you to find the number of *milligrams* of sodium chloride present in

#color(blue)("1 g" = 10^3"mg") " "# and#" " color(blue)("1 L" = 10^3"mL")#

So, if your solution contains **per liter**, you can say that it contains

#6.277 "g"/color(red)(cancel(color(black)("L"))) * (1 color(red)(cancel(color(black)("L"))))/(10^3"mL") = 6.277 * 10^(_3)"g/mL"#

This will be equivalent to

#6.277 * color(purple)(cancel(color(black)(10^(-3)))) color(red)(cancel(color(black)("g")))/"mL" * (color(purple)(cancel(color(black)(10^3)))"mg")/(1color(red)(cancel(color(black)("g")))) = "6.277 mg/mL"#

You now know that

#"6.277 g/L " = " 6.277 mg/mL"#

So, if

#15.0 color(red)(cancel(color(black)("mL solution"))) * overbrace( ("6.277 mg NaCl")/(1color(red)(cancel(color(black)("mL solution")))))^(color(blue)("the given concentration")) = "94.155 mg NaCl"#

Rounded to three sig figs, the number of sig figs you have for the volume of the sample, the answer will be

#m_(NaCl) = color(green)("94.2 mg NaCl")#