# 59.0 mL of a 1.60 M solution is diluted to a total volume of 268 mL. A 134-mL portion of that...

## Question:

59.0 mL of a 1.60 M solution is diluted to a total volume of 268 mL. A 134-mL portion of that solution is diluted by adding 173 mL of water. What is the final concentration? Assume the volumes are additive.

## Calculation of the final concentration of the solution that has undergone dilution for 3 times from the stock solution.

The concentration of any stock solution is decreased by diluting with a suitable solvent, water is frequently used. However, it is more advisable to use the same solvent that is used in the preparation of the stock solution. We will stick to the principle of conservation of molecules or ions in the solutions and come to the final conclusion.

Since the dilution is not in multiples, it is better to choose the quantity of mmol present in the solution and at each stage.

59 ml of 1.6 M have 59*1.6 = 94.4 mmol

Since 59 ml is diluted to the total volume of 268 ml,

268 ml contains 94.4 mmol

∴ 134 ml contains 47.2 mmol

Now, 134 ml is diluted by adding 173 ml of water, so the total volume is 307 ml

307 ml contains 47.2 mmol

∴ 1000 ml contains {eq}\frac{1000 * 47.2}{307} = 153.7459 mmol \quad or \quad 0.1538 M {/eq}

The final concentration is 0.1538 M