A 71.0 mL 71.0 mL aliquot of a 1.30 M 1.30 M solution is diluted to a total volume of 248 mL. 248 mL. A 124 mL 124 mL portion of that solution is diluted by adding 133 mL 133 mL of water. What is the final concentration? Assume the volumes are additive.

The answr to the question is

The final concentration = 0.179 M

Explanation:

To solve the question, we note the dilution factor as follows

The dilution factor is given by

Therefore we have, for the first dilution

71 mL of 1.3 M diluted to 248 mL which gives

= 1.3×71/248 = 0.372 M

to the 124 mL portion of the diluted solution 133 mL of water is added thus we have 124 +133 =257 mL

We therrefore proceed as folows

where = concentration of new solution

= 0.372 M

= 124 mL

= 257 mL

Therefore   = 0.372×124/257 = 0.179 M

The final concentration = 0.179 M

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