Question

Calculate the following quantity: volume of 1.403 M copper(II) nitrate that must be diluted with water to prepare 575.2 mL of a 0.8012 M solution.

Answer 1

Answer:

328 ml

Explanation:

We have given final volume =575.2 ml=0.575 L

Final concentration = 0.8012 M

We know that moles of copper(II) nitrate = final volume ×final concentration =0.8012×.0575=0.4606 moles

We have to find initial volume

So initial volume $=\frac{moles\ of \ copper(II)\ nitrate}{initial\ concentration}=\frac{0.4606}{1.403}=0.328L=328ml$