Question

Question #0db2c

Answer 1

`"8500 ppm"`
`"0.15 M"`

Explanation:

You're dealing with a `0.85%` percent concentration by mass, or `"%w/w"`, sodium chloride solution.

A solution's percent concentration by mass is defined as the mass of the solute, which in your case is sodium chloride, divided by the total mass of the solution, and multiplied by `100`.

`color(blue)("%w/w" = "mass of solute"/"mass of solution" xx 100)`

In simple terms, a solution's percent concentration by mass tells you how many grams of solute you get per `"100 g"` of solution.

In your case, `0.85%"w/w"` sodium chloride solution will have `"0.85 g"` of sodium chloride for every `"100 g"` of solution.

To get a solution's concentration in parts per million, or ppm, you need to divide the mass of the solute, expressed in grams, by the mass of the solution, expressed in grams, and multiply the result by `10^6`.

`color(blue)("ppm" = "mass of solute"/"mass of solution" xx 10^6)`

To make calculations easier, let's take a `"100-g"` sample of this saline solution. This sample will contain `"0.85 g"` of sodium chloride, which means that its concentration in ppm will be

`(0.85color(red)(cancel(color(black)("g"))))/(100color(red)(cancel(color(black)("g")))) xx 10^6 = color(green)("8500 ppm")`

To get the concentration in moles per liter, you need to use sodium chloride's molar mass to determine how many moles you get in `"0.85 g"`

`0.85color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.443color(red)(cancel(color(black)("g")))) = "0.01454 moles NaCl"`

To get the volume of the solution, you need to know its density.

http://www.maelabs.ucsd.edu/mae171/Conc%20vs%20density.pdf

SInce the solution's percent concentration by mass is so small, you can safely assume that the density of the solution wil be equal to that of water at room temperature, which is `~~ 10^3"g/L"`.

The volume of the solution - expressed in lites - will be

`100color(red)(cancel(color(black)("g"))) * "1 L"/(10^3color(red)(cancel(color(black)("g")))) = "0.1 L"`

Therefore, the solution's molarity will be

`color(blue)(c = n/V)`

`c = "0.01454 moles"/"0.1 L" = color(green)("0.15 M") ->` rounded to two sig figs