{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chp2_MixDilute

# Chp2_MixDilute - Mixing and Dilution Consider an initial...

This preview shows pages 1–2. Sign up to view the full content.

Mixing and Dilution Consider an initial solution with molar concentration, C i , and volume, V i . Note that the number of mols of solute is thereby: C i × V i = n i . To this solution, let us imagine adding a volume, V 2 , of a second (of the same type) characterized by a concentration, C 2 . Again, note the number of moles added to the original solution is n 2 = C 2 × V 2 . The resulting, “final”, solution, will have a concentration, C f and volume, V f . The basic rule is that the final concentration of any solution formed by mixing two (or more) solutions is: “the total amount of solute divided by the total amount of solvent/solution.” The solvent/solution distinction depends on the particular concentration unit. For molarity, the denominator would be the total volume of solution. Thus, we may write: 2 2 2 2 2 V V V C V C V V n n C i i i i i f + + = + + = This expression is explicit for mixing two solutions. More generally for the case of mixing “N” different solutions: + + + + + + = = = = 3 2 1 3 3 2 2 1 1 1 1 V V V V C V C V C V V C C N i i N i i i f For those of you with some statistics, the above equation is simply the weighted average formula.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 2

Chp2_MixDilute - Mixing and Dilution Consider an initial...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online