Question

A 75.0 mL 75.0 mL aliquot of a 1.70 M 1.70 M solution is diluted to a total volume of 278 mL. 278 mL. A 139 mL 139 mL portion of that solution is diluted by adding 165 mL 165 mL of water. What is the final concentration? Assume the volumes are additive.

Answer 1

Answer:

0.210 M

Explanation:

A 75.0 mL aliquot of a 1.70 M solution is diluted to a total volume of 278 mL.

In order to find out the resulting concentration (C₂) we will use the dilution rule.

C₁ × V₁ = C₂ × V₂

1.70 M × 75.0 mL = C₂ × 278 mL

C₂ = 0.459 M

A 139 mL portion of that solution is diluted by adding 165 mL of water. What is the final concentration? Assume the volumes are additive.

Since the volumes are additive, the final volume V₂ is 139 mL + 165 mL = 304 mL. Next, we can use the dilution rule.

C₁ × V₁ = C₂ × V₂

0.459 M × 139 mL = C₂ × 304 mL

C₂ = 0.210 M

Answer 2

Answer:

The correct answer is 0.21 M

Explanation:

We have an initial solution of concentration 1.70 M and we dilute it twice. In each dilution we can calculate the final concentration (Cf) from the initial concentration (Ci) and the final and initial volumes respectively (Vf and Vi) as follows:

Cf x Vf= Ci x Vi

Cf= Ci x Vi/ Vf

In the first dilution, Ci is 1.70 M, the initial volume we take is 75 ml and the final volume is 278 ml.

Cf= 1.70 M x 75 ml / 278 ml = 0.46 M

In the second dilution, the initial concentration is the previously obtained (Ci= 0.46 M), the initial volume is Vi= 139 ml and the final volume is the addition of 139 ml and 165 ml (because we add 165 ml to 139 ml). Thus:

Cf= (0.46 M x 139 ml)/ (139 ml + 165 ml) = 0.21 M