August 22, 2019, 01:34:58 AM
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Topic: Amount of solute in a Molarity Concentration.  (Read 114 times)

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Offline Greek0000

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Amount of solute in a Molarity Concentration.
« on: August 11, 2019, 11:40:36 AM »
I am trying to solve for grams in solute of lactic acid concentration of 0.10 M C3H6O2

notice the example and my answer was 7.4 grams of lactic acid (see attachment)

Offline Borek

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Re: Amount of solute in a Molarity Concentration.
« Reply #1 on: August 11, 2019, 11:58:24 AM »
Pure nonsense - you can't speak about amount of substance without knowing volume of the solution. Throw away this book (or whatever it is).
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Offline sunkal

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Re: Amount of solute in a Molarity Concentration.
« Reply #2 on: August 11, 2019, 04:24:12 PM »
Greek0000,

The setup shown in the attachment under NOTE inside the box is: 0.10 M C3H6O2 x 74 g mol C3H6O2 / 1 mol C3H6O2. You should realize that this happens to be a totally wrong setup. If you do what is called dimensional analysis, you can realize it.

First, notice that mol (in 74 g mol) and mol (in 1 mol) will cancel leaving g. M stands for mol / L. So, the final unit left is g mol / L. It is not just g, as it says in the final answer.

Secondly, the correct setup for the conversion factor is: 74 g C3H6O2 / 1 mol C3H6O2. This comes from the molar mass. 74 g mol in the numerator is completely wrong.

Multiplying the molarity by molar mass will never give any useful quantity. One needs to know the volume in L. Only then a suitable dimensional analysis will occur. The equation to be used is n = M V and then the resulting n can be converted to g using the molar mass.

This is why Borek refers to it as pure nonsense.

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