Question

A chemist adds 135.0 mL of a 0.21M zinc nitrate (Zn(NO3) solution to a reaction flask. Calculate the mass in grams of zinc nitrate the chemist has added to the flask. Be sure your answer has the correct number of significant digits.

Answer 1

Answer:

5.36 grams the mass in grams of zinc nitrate the chemist has added to the flask.

Explanation:

$Molarity=\frac{\text{Moles of solute}}{\text{Volume of the solution (L)}}$

Moles of zinc nitrate = n

Volume of the solution = 135.0 mL = 0.1350 L

Molarity of the solution = 0.21 M

$0.21 M=\frac{n}{0.1350 L}$

$n=0.21M\times 0.1350 L=0.02835 mol$

Mass of 0.02835 moles of zinc nitrate:

0.02835 mol × 189 g/mol = 5.358 g ≈ 5.36 g

5.36 grams the mass in grams of zinc nitrate the chemist has added to the flask.