FL9 #9 bio/biochem section

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FL9 #9 bio/biochem section
I am still confuse by this problem. How did you get "dissolved particles by a factor of 1.5"? I get 2 for the 1 M KCl (split K and Cl). Am I missing something?

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Re: FL9 #9 bio/biochem section
Hi jumpinfeet,
Very good question! I think you might be missing the need to account for the volume of the KCl solution  the question says that you add "an equal volume of 1 M KCl solution", not that you dissolve KCl in the original solution until you reach a concentration of 1 M. This means that the new solution has double the volume of the original one and three times the concentration of solvents. It may be helpful to work through this with easy (although unrealistically large) numbers  let's say that the original solution had 1 mole of glucose in 1 L solvent, so the combined molarity (iM) = 1 M. In the new solution, we will have 3 moles of solute (1 mol K+, 1 mol Cl, 1 mol glucose) in 2 L of solvent, so the combined molarity (iM) = 3/2 = 1.5 M. Hence the factor of 1.5.
Hope this helps, and feel free to reach out w/ further followup!
Very good question! I think you might be missing the need to account for the volume of the KCl solution  the question says that you add "an equal volume of 1 M KCl solution", not that you dissolve KCl in the original solution until you reach a concentration of 1 M. This means that the new solution has double the volume of the original one and three times the concentration of solvents. It may be helpful to work through this with easy (although unrealistically large) numbers  let's say that the original solution had 1 mole of glucose in 1 L solvent, so the combined molarity (iM) = 1 M. In the new solution, we will have 3 moles of solute (1 mol K+, 1 mol Cl, 1 mol glucose) in 2 L of solvent, so the combined molarity (iM) = 3/2 = 1.5 M. Hence the factor of 1.5.
Hope this helps, and feel free to reach out w/ further followup!
Andrew D.
Content Manager, Next Step Test Prep.
Content Manager, Next Step Test Prep.
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