Question

62.0 mL of a 1.50 M solution is diluted to a total volume of 258 mL. A 129-mL portion of that solution is diluted by adding 187 mL of water. What is the final concentration? Assume the volumes are additive.

Answer 1

Moles of compound in the original solution=molarity* volume of solution=1.50M*62.0ml=1.50 mol/L *62.0/1000 L=0.093 moles

Let Concentration of original solution=C1=1.50 M

Volume of original solution=V1=62 ML

Concentration after first dilution=C2=?

Volume after first dilution=V2=258 ML

Use equation,

C1*V1=C2*V2

C2=C1*V1/V2=1.50 M*62.0 ml/258 ml=0.360 M

Concentration after second dilution=C3=?

Volume after second dilution=V3=187 ml

C3*V3=C2*V2

C3=final concentration =C2*V2/V3=0.360m*129 ML/187 ml=0.2483 M (answer)