Michelle Fu 1H
Posts: 35
Joined: Fri Sep 28, 2018 12:19 am

In class Dr. Lavelle had a question that went:

What volume of a .0380 M KMnO4 is needed to prepare 250 mL of 1.5 x 10^(-3) MKMnO4?

While I understand the math and the answer (9.87 x 10^(-3) L or 9.8 mL, I'm wondering how conceptually this works out as we start off with a more concentrated solution with less water and it somehow prepares into more solution that's less concentrated. I think I'm misinterpreting part of the problem through the word "prepare" but I'm not sure.

Andrew Sun 1I
Posts: 30
Joined: Fri Sep 28, 2018 12:18 am

### Re: Confused about molarity concept

i think it means you need to mix in 9.8ml (the answer) of the 0.0380M KmNO4 into say a volumetric flask, then fill it up to 0.250L with water, and that will create the 1.50x10^-3M solution of KmNO4. At least I think that is how it can be interpreted, open to other thoughts

Jonathan Pai 2I
Posts: 121
Joined: Fri Sep 28, 2018 12:27 am

### Re: Confused about molarity concept

Calculation aside, M(i)V(i)=M(f)V(f) answers your question. Conceptually, knowing that the amount of solute remains the same, if you add more water (Volume is increasing), then even though the amount of solute is the same, it is less concentrated because it is now in a larger area/container/space (Concentration is decreasing). From a mathematical perspective, because it is an equation, adding more water (diluting) increases V(f), therefore M(f) must decrease to keep the equation equal to M(i)*V(i).

AngelaZ 1J
Posts: 65
Joined: Fri Sep 28, 2018 12:23 am

### Re: Confused about molarity concept

As Andrew said, 9.8 mL of KMnO4 is poured into a beaker and water is added up to the 250 mL mark. So, the stock solution is being diluted with water, and the concentration of the solution is decreased since there is now the same amount of solute in more water.

taryn_baldus2E
Posts: 62
Joined: Fri Sep 28, 2018 12:24 am

### Re: Confused about molarity concept

Seeing as the solute remains the same, by adding more water you are decreasing the molarity of the solution. If you refer to the equation Molarity = moles of the solute/volume of the solution, you can apply it to the example from the lecture by acknowledging that though the solute does not change, the volume of the solution will increase. Since the volume acts as the denominator of this equation, as the volume increases, the product (also known as the molarity of the solution) will decrease, indicating that the two measurements are inversely proportional.