# 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?

Jan 22, 2016

$\text{94.2 mg}$

#### 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 $\text{15.0 mL}$ of this solution, provided that you use the conversion factors

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

So, if your solution contains $\text{6.277 g}$ of sodium chloride per liter, you can say that it contains

$6.277 \text{g"/color(red)(cancel(color(black)("L"))) * (1 color(red)(cancel(color(black)("L"))))/(10^3"mL") = 6.277 * 10^(_3)"g/mL}$

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

$\text{6.277 g/L " = " 6.277 mg/mL}$

So, if $\text{1 mL}$ of this solution contains $\text{6.277 mg}$ of sodium chloride, $\text{15.0 mL}$ will contain

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

${m}_{N a C l} = \textcolor{g r e e n}{\text{94.2 mg NaCl}}$