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QUESTION

# 1.7 g of NaNO3 (Mr = 85) is dissolved in water to prepare 0.20 dm3 of solution. What is the concentration of the resulting solution in mol dm–3?

"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

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