Question

A 75.0 mL aliquot of a 1.50 M solution is diluted to a total volume of 208 mL. A 104 mL portion of that solution is diluted by adding 185 mL of water. What is the final concentration? Assume the volumes are additive. concentration: M

Answer 1

Solution:

First calculate moles from 75.0mL and 1.50 M:

75.0 mL x 1.50 M =112.5 mmol

This mmol diluted to a total volume of 208 mL

112.5 mmol / 208 mL = 0.5409 M

After that we take 104 mL above solution-

0.5409 M x 104 mL = 56.23 mmol

After that in this solution added 185 mL water

Total volume = 185 + 104 = 289 mL

56.23 mmol/289 mL = 0.1946 M

Hence the final concentration of the solution is =0.1946 M