The given question is incomplete. The complete question is:
A chemist adds 35.0 mL of a 2.82M sodium nitrate (NaNO3) solution to a reaction flask. Calculate the millimoles of sodium nitrate the chemist has added to the reaction flask.
Answer: 98.7 mmol
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
[tex]Molarity=\frac{n}{V_s}[/tex]
where,
n = milli moles of solute
[tex]V_s[/tex] = volume of solution in ml
Now put all the given values in the formula of molarity, we get
[tex]2.82=\frac{n}{35.0}[/tex]
[tex]n=2.82\times 35.0=98.7mmol[/tex]
Therefore, the millimoles of sodium nitrate the chemist has added to reaction flask are 98.7
