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

## 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?

## 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.

## The Attempt at a Solution

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symbolipoint
Homework Helper
Gold Member
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
Homework Helper
Gold Member
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.

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?

Yes? No?

symbolipoint
Homework Helper