Question

A chemist adds of a M sodium nitrate solution to a reaction flask. Calculate the millimoles of sodium nitrate the chemist has added to

Answer 1

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

Step-by-step explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

`Molarity=(n)/(V_s)`

where,

n = milli moles of solute

`V_s` = volume of solution in ml

Now put all the given values in the formula of molarity, we get

`2.82=(n)/(35.0)`

`n=2.82* 35.0=98.7mmol`

Therefore, the millimoles of sodium nitrate the chemist has added to reaction flask are 98.7