A chemist adds of a M sodium nitrate solution to

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kcarstensen59070

17.03.2022

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.

98.7 mmol

Explanation:

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

$Molarity=\frac{n}{V_s}$

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

$2.82=\frac{n}{35.0}$

$n=2.82\times 35.0=98.7mmol$

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

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

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