The question is incomplete. The complete question is:
A chemist adds 390.0 mL of a 2.1M sodium nitrate [tex]NaNO_3[/tex] solution to a reaction flask. Calculate the mass in grams of sodium nitrate the chemist has added to the flask. Round your answer to 2 significant digits.
Answer: 70 grams
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]
where,
n = moles of solute
[tex]V_s[/tex] = volume of solution in ml
moles of [tex]NaNO_3[/tex] = [tex]\frac{\text {given mass}}{\text {Molar Mass}}=\frac{xg}{85g/mol}[/tex]
Now put all the given values in the formula of molality, we get
[tex]2.1M=\frac{x\times 1000}{85\times 390.0}[/tex]
[tex]x=70g[/tex]
Therefore, the mass in grams of sodium nitrate the chemist has added to the flask is 70 grams