Formula M1V1=M2V2  [ENDORSED]

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Jasmin Argueta 1K
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Formula M1V1=M2V2

Postby Jasmin Argueta 1K » Thu Oct 11, 2018 10:11 pm

The formula M1V1=M2V2 was not in the text, is that correct? For problem G13 it seems much easier to use that formula but shows another way in the text. We just know to use it when we are given say, 2 volumes and one molarity for instance?

Chem_Mod
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Re: Formula M1V1=M2V2

Postby Chem_Mod » Thu Oct 11, 2018 10:16 pm

That equation was used, but the solutions manual does not show it (the fact that the question says to solve without a calculator means the calculation could be done mentally).

Nicholas Carpo 1L
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Re: Formula M1V1=M2V2  [ENDORSED]

Postby Nicholas Carpo 1L » Thu Oct 11, 2018 10:17 pm

In G13, you still use that formula to some extent to solve the molarity of NH4NO3, although the calculation is not shown in the solution manual.

The equation M1V1=M2V2 becomes (0.20 M) (1.0 L) = (x) (1.0 + 3.0 L) which leads to the knowledge that the molarity of the diluted solution is 0.050 M.

You would then need to double this number as there are two N moles per mole of NH4NO3.

Rachel Dang 1H
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Re: Formula M1V1=M2V2

Postby Rachel Dang 1H » Thu Oct 11, 2018 10:19 pm

Yes we can use the that formula whenever we are given 3 of the 4 values or information to find those values.

Kayla Vo 1B
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Joined: Fri Sep 28, 2018 12:26 am

Re: Formula M1V1=M2V2

Postby Kayla Vo 1B » Thu Oct 11, 2018 10:36 pm

Yes this equation can be used to solve problems like dilutions. Just be sure that your units always are converted to liters and mols/L (M) so that they cancel out during calculations.

SophiaKohlhoff4B
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Re: Formula M1V1=M2V2

Postby SophiaKohlhoff4B » Sun Oct 14, 2018 9:41 pm

This formula is especially useful in diluting stock solutions.


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