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

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Discussion Overview

The discussion revolves around the preparation of a 250 mL solution of 0.180 M K2S2O8, focusing on the correct calculation of the mass of K2S2O8 needed and the procedural steps for preparing the solution using solid K2S2O8, deionized water, a balance, and a volumetric flask.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant initially calculates the mass of K2S2O8 needed as 6.66 x 10^-4 g, but does not correctly account for the volume of the solution.
  • Another participant points out errors in the mass calculation, emphasizing the need to calculate for 250 mL instead of 1 L and to ensure correct unit consistency.
  • A subsequent reply reiterates the importance of correctly interpreting the molarity definition and recalculates the mass to 12.16 g of K2S2O8.
  • Further suggestions include ensuring that all material is quantitatively transferred to the flask and properly mixed with deionized water.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the initial mass calculation, with corrections and recalculations being made. There is a general agreement on the final mass of 12.16 g, but the discussion reflects uncertainty in the initial steps.

Contextual Notes

Limitations include potential misunderstandings of molarity and volume conversions, as well as the need for clarity in unit conversions throughout the calculations.

Who May Find This Useful

This discussion may be useful for students preparing solutions in chemistry labs, particularly those learning about molarity and solution preparation techniques.

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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.

 
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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)
 
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?
 
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.
 

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