Question #7cad8

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