Lab Preparation Using a solid, DI water, a balance, and a flask

[TypeFun]

Homework Statement

Describe how you would prepare exactly 250 mL of .180 M K₂S₂O₈ solution for use. Assume that you have a supply of solid K₂S₂O₈, deionized water, a balance, and a volumetric flask. Also, how many grams of K₂S₂O₈ would you need?

Homework Equations

The Attempt at a Solution

.180 M K₂S₂O₈ / 270.32 g/mol = 6.66 x 10^-4 g K₂S₂O₈

Pour 250 mL of DI water into a volumetric flask. Balance out 6.66 x 10^-4 g of K₂S₂O₈. Add the two together.

 

[symbolipoint]

Check the units for your mass calculation and you will find that they are wrong. Your units should, on both sides of the equation, be "grams".

Assuming your molecular weight of the compound K₂S₂O₈ is correct, your mass calculation should be not for 1 L of solution, but for 250 ml. of solution, meaning you want something like:
250 * 10^-3L * 0.180 (moles / L) * 270.32 (grams / mole)

 

[chemisttree]

Nope. Try again. Remember that 0.180M is 0.180 moles in one liter, not 1/4 liter.

Also remember to set up the problem so that you get the units you want. You set it up incorrectly. 6.66 X 10^-4 g is pretty wrong.

 

[TypeFun]

Hey thanks. My new number is 12.16 g of K2S2O8. Would that change my answer to: Using the balance, measure out 12.16 g of K2S2O8. Add that to a volumetric flask. Pour DI water into the flask, until 250 mL has been reached.

Does that sound correct?

 

[TypeFun]

Yes? No?

 

[symbolipoint]

Good.
You may well wish to diligently transfer all the weighed material quantitatively to the flask, using squirt of deionized water from squirt bottle, and enough water to dissolve the material, and then fill to the 250 ml. mark, stopper and invert a few times to finish mixing.