Question

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.

Answer 1

Answer:

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

$DF = \frac{V_{Diluted} }{V_{Concentrated} } = \frac{C_{Concentrated} }{C_{Diluted} }$

Therefore we have, for the first dilution

71 mL of 1.3 M diluted to 248 mL which gives

$C_{Diluted} = \frac{C_{Concentrated}*V_{Concentrated} }{V_{Diluted}}$ = 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

$C_{Diluted} = \frac{C_{Concentrated}*V_{Concentrated} }{V_{Diluted}}$ where $C_{Diluted}$ = concentration of new solution

$C_{Concentrated}$ = 0.372 M

$V_{Concentrated}$ = 124 mL

$V_{Diluted}$ = 257 mL

Therefore  $C_{Diluted}$ = 0.372×124/257 = 0.179 M

The final concentration = 0.179 M