Question

Question #7cad8

Answer 1

`"0.10 mol dm"^(-3)`

Explanation:

Your strategy here will be to use the molar mass of the compound and the sample given to you to find how many moles of sodium nitrate, `"NaNO"_3`, you will have in solution.

Once you know how many moles of you have, use the volume of the solution to calculate its molarity.

So, a substance's molar mass tells you the mass of one mole of that substance. In your case, sodium nitrate is said to have a molar mass of `"85 g/mol"`, which means that every mole of this compound has a mass of `"85 g"`.

This means that your `"1.7-g"` sample will contain

`1.7 color(red)(cancel(color(black)("g"))) * "1 mole NaNO"_3/(85 color(red)(cancel(color(black)("g")))) = "0.020 moles NaNO"_3`

Now, it's important to realize that

`"1 dm"^3 = "1 L"`

This means that your solution will have a volume of `"0.20 L"`.

As you know, molarity is defined as moles of solute, which in your case is sodium nitrate, divided by liters of solution.

`color(blue)("molarity" = "moles of solute"/"liters of solution")`

Since you have everything that you need, plug your values into this equation and find the molarity of the solution

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

`c = "0.020 moles"/"0.20 L" = "0.10 mol L"^(-1) = "0.10 mol dm"^(-3) = "0.10 M"`

SIDE NOTE As a fun fact, your solution will not actually be
`"0.10 M"` sodium nitrate because sodium nitrate dissociates completely in aqueous solution to form

`"NaNO"_text(3(aq]) -> "Na"_text((aq])^(+) + "NO"_text(3(aq])^(-)`

Since you have `1:1` mole ratios between all three chemical species, you can say that your solution will be `"0.10 M"` in `"Na"^(+)` and `"0.10 M"` in `"NO"_3^(-)`.

If you want, you can read more on this concept here

https://socratic.org/questions/what-is-formality-explain-with-an-example