##”0.10 mol dm”^(-3)##
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 .
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, is defined as moles of , 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## 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
