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o prepare a 200 mL of aqueous stock solution X, you transferred 50 mL of a known concentration solution into a 200 mL volumetri...Asked by Brad
1) To prepare a 200 mL of aqueous stock solution X, you transferred 50 mL of a known concentration solution into a 200 mL volumetric flask. If you realize that the volumetric flask was contaminated with water droplets after you cleanly transferred the known concentration solution. The 200 mL solution X will have a
_________(higher, lower or same) concentration than(as) you originally calculated.
2) You transfer a 2M stock solution of Y into a container using a slightly wet pipet containing water drops. The concentration of the transferred solution Y has a
___________(lower, higher, or same) concentration than(as) 2M
_________(higher, lower or same) concentration than(as) you originally calculated.
2) You transfer a 2M stock solution of Y into a container using a slightly wet pipet containing water drops. The concentration of the transferred solution Y has a
___________(lower, higher, or same) concentration than(as) 2M
Answers
Answered by
bobpursley
why would water dropplets matter, you are filling it up to 200ml with water added.
If you add water to the 2M solution, is is diluted already before you mix.
If you add water to the 2M solution, is is diluted already before you mix.
Answered by
Brad
1) So since we are just filling it up to 200ml the volume does not change so it has the SAME concentration?
2) so if we add more water which increases the volume making it (mol / L <-- increased) resulting in a LOWER concentration?
2) so if we add more water which increases the volume making it (mol / L <-- increased) resulting in a LOWER concentration?
Answered by
DrBob222
right on both.
For the second one you can also do it another way. Filling a WET pipet with the solution means you have LESS of the 2 M solution and M = mols/L; therefore, we have fewer mols and that means smaller M.(lower concn).
For the second one you can also do it another way. Filling a WET pipet with the solution means you have LESS of the 2 M solution and M = mols/L; therefore, we have fewer mols and that means smaller M.(lower concn).
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