When dealing with molar concentrations, you might be asked to give directions. In cases like this, you usually have to find the products of solution and figure out how much mass you have to add to the solution.

• if you add water the number of moles in the solution stay the same
• if the volume changes the solution will became less concentrated

ex.

Ello, I am Ron Weasley. I was asked to make me a 500 mL solution of MgSO4 with a concentration of 0.98M, bloody hell if I do say so meself. Bugger, what steps should I follow?

1. Determine the number of moles by dividing the molarity by volume
   
   • V=500mL-->0.5L / o.98mol/L x 0.50L=0.49 mol
2. After finding the number of moles, determine what the mass of MgS
   
   • Mg=1(24.3), S=1(32.1), O=4(16) }120.4 g/mol
3. Multiply moles by grams to find the overall mass
   
   • 0.49mol x 120.4g = 58.99 g
4. After finding the mass, write the follow up steps for making the solution ex:
   
   • weigh 59 g of MgSO4
   • measure 500 mL of water
   • stir solution

Dilution of Solution

• when you add water, the concentration decreases
• if volume is doubled then the concentration is halved

volume con'c moles
6.0 L 2.0 mol/L 12.0 mol
12.0 L 1.0 mol/L 12.0 mol


example:

I am Optimus Prime and I added 300 mL of water to 60 mL of 0.90M of HF. What is the final [HF]

1. In this question, it's best to use: C1V1=C2V2
2. C1 and V2 are the first concentration and first volume
3. C2 and V2 are the final concentration and volume
   
   • C1= 0.90/ V1=300mL or 0.3L / C2=unknown/ V2=360 mL or 0.36 L
   • (0.90)(0.3)=C2(0.36)/(0.36)
   • 0.27/0.36=C2
   • C2=0.75 M

Here is a video explaining concentration a little more. ..


Please excuse the examples.. haha

Hi, Did you know that the test of all knowledge is experiment. :)

Chemistry 11, Block E
Mr. Doktor

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