# If 92.8 grams of sodium nitrate are dissolved in 2.3 L of water, what is the resulting solution concentration?

Update:

Can someone explain how to do this please?

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• With only the information provided you cannot calculate the molaRity accurately.

92.8 g of NaNO3 is a significant fraction of 2.3 kg of water, so you cannot assume that the addition of the NaNO3 does not change the volume of water. The missing information is the density of the solution after the NaNO3 has been dissolved.

What you can calculate with what's given is the molaLity:

(92.8 g NaNO3) / (84.9947 g NaNO3/mol) / (2.3 kg H2O) = 0.475 mol/kg =

0.475 m

• d3 weeks agoReport

Thank you. That clears some things up for me.

• find the molar mass of NaNO3 ... 85g/mole

92.8g / 85 = 1.092 mole

1.092 / 2.3 = 0.475 M (molarity) <<< answer

• david
Lv 7
3 weeks agoReport

There are several different ways to calculate concentration. The problem usually specifies what king of concentration .. %by mass, or molarity (M), or molality (m) ... but you did not specify which, so I just guessed that you wanted molarity ... it is moles solute / liters of solution....