Question

How many milliliters are required to prepare 100.0 mL of a solution that has 5.0 mg Cu/mL from a solution of Cu(NO₃)₂ 0.80 M? The answer is 9.8 mL, can someone please explain how to get this result?

Answer 1

One molecule of Cu(NO₃)₂ contain 1 Cu atom. Therefore, we can write , moles of Cu(NO₃)₂ = moles of Cu or

[ Cu(NO₃)₂ ] = [ Cu ] = 0.80 M

We know that, Molarity = No. of moles of solute / volume of solution in L

On rearranging, we get No. of moles of solute = Molarity volume of solution in L

Assume we have 1 L solution of Cu(NO₃)₂ then ,

No. of moles of Cu = 0.80 mol / L 1 L = 0.80 mol

We have, no. of moles = Mass / molar mass

Mass of Cu = No. of moles of Cu   Molar mass of Cu

Mass of Cu = 0.80 mol 63.54 g / mol

Mass of Cu = 50.8 g

i e 1 L solution of 0.80 M Cu(NO₃)₂ contain 50.8 g Cu.

Now, we can calculate above concentration into mg / ml.

1000 ml solution 50800 mg Cu   

1 ml solution 50800 / 1000 mg Cu

1 ml solution 50.8 mg Cu

i e we get Concentration of Cu solution = 0.80 M = 50.8 mg / ml

Now, we can apply dilution formula to find out volume of 50.8 mg Cu / ml solution required to prepare 5 mg Cu / ml solution.

we have dilution formula, C _stock V _stock = C _dilute V _dilute

V _stock = C _dilute V _dilute / C _stock

V _stock = 5.0 mg / ml 100.0 ml / 50.8 mg / ml

V _stock = 9.8 ml