Question

(4 pts) Determine the volume of 0.75 M stock solution needed to prepare 50.0 mL of a 0.15 M solution. Show the complete calculation with units. *Note ‐ If you struggle with this calculation, refer to experiment 5.

(8 pts) Write out a complete, detailed experimental procedure that you (or anyone else) could follow to carry out the solution preparation you have described above. Your procedure must follow these restrictions and guidelines:

 You must prepare the 50.0 mL in one container.

 You must use the most precise glassware possible out of the options discussed in the previous

lab.

 Your procedure must include detailed information about the exact size of equipment to use

(e.g. 5.0 mL pipet vs 25.0 mL pipet, etc.) as well as the exact volumes/quantities to be

measured.

 The steps in your procedure should be detailed enough that another student could follow the

procedure’s instructions to complete the experiment, even if the student had never previously seen the included experiment or activities.

(4 pts) If a sample started as a mixture of powder rather than a solution, write the additional steps that would be needed to prepare a 0.75M stock solution from solid powder. Be sure your expanded procedure includes the same detailed information about exact size of equipment and any quantity information (mass/volume/etc.) that you would need to be measured to make an accurate and precise stock solution.

Answer 1

ANSWER :

If a sample started as a mixture of powder rather than a solution, write the additional steps that would be needed to prepare a 0.75M stock solution from solid powder. information about exact size of equipment and any quantity information (mass/volume/etc.)

Assuming that the solution is of a compound X which has MW of 'x', the preparation is as follows:

In order to prepare a stock solution of this compound of conc 0.75 M, and assuming that you want to prepare 250 mL of this stock solution:

Moles of compound required = Molarity*Volume = 0.75*0.25 = 0.1875

So, mass of compound needed = 0.1875*x

So, dissolve this much amount of the powder in 250 mL distilled water and this would give you a stock solution of volume 250mL having conc. 0.75 M

Next, as you said, if you take 10 mL of this stock solution, and make the final volume upto 50 mL, this corresponds to a 5 times dilution.

So the conc of the dilution is: 0.75/5 = 0.15 M