Chemistry: Calculating Molarity II
In an earlier post I talked about what a mole is in chemistry. You need to understand what a mole of substance is before you can understand concentrations and how to make up a molar solution.
A molar solution is defined as a mole of substance dissolved in 1 litre (or liter) of solute.
Now this is not strictly true as a liter is not really a correct unit and a molar solution should be defined as a mole of substance dissolved in 1 dm3 (1 deci-meter-cubed) of solute. (1 dm3 pretty much interchangeable, but in fact 1 dm3 is 1,000 cm3, whereas 1 liter is 1,000.028 cm3). Anyway, I digress….
So, what is this solute. Well, it can be any liquid. Water, an acid, an alcohol, anything. It is just a liquid that is used to dissolve the solid.
As I said in the previous post, 1 mole of a substance contains 6.0022 x 1023 atoms, ions or molecules. And if you dissolve a mole of a substance in a liter of solute (water, acid, alcohol etc.) you have a 1 Molar (or 1 M) solution. This is it. That is all you have to do.
The equation that can be used to work out the molarity of a solution is:
| Where: | ||
| M | = | Concentration (M) |
| W | = | Weight (mass) (g) |
| MW | = | Molecular weight |
| V | = | Volume (l) |
So, for example: What is the molarity of a 5 g of a compound of molecular weight 50, dissolved in 250 ml? Using the above equation:
| Where: | |||
| M | = | Conc. (M) | = ? |
| W | = | Weight (g) | = 5 |
| MW | = | Molecular weight | = 50 |
| V | = | Volume (l) | = 0.25 l |
∴ M = (5 ÷ 50) ÷ 0.25
∴ M = 0.1 ÷ 0.25 = 0.4 M
So, 5 g of a compound of molecular weight 50, dissolved in 250 ml makes a 0.4 M solution.
(Another way of thinking of this is you have 5 g of the compound in 250 ml, you have 5/250 g/ml = 0.02 g/ml. Multiply by 1000 to give you grams per liter, so, 0.02 x 1000 = 20 g/l. 20 ÷ MW is the concentration, ∴ 20 ÷ 50 = 0.4 M.)
The number of grams required to make up a solution is calculated using:
| Where: | ||
| M | = | Concentration (M) |
| W | = | Weight (mass) (g) |
| MW | = | Molecular weight |
| V | = | Volume (l) |
(This above equation is just the first one re-arranged. It is not really a different equation.)
So, for example: How many grams of compound with a molecular weight of 75 are needed to make 75 ml of a 0.1 M solution? Using the above equation:
| Where: | |||
| M | = | Conc. (M) | = 0.1 M |
| W | = | Weight (g) | = ? |
| MW | = | Molecular weight | = 75 |
| V | = | Volume (l) | = 75 ml |
So, get the concentrations and volumes to ‘base’ units of Molar and liters (litres). Therefore, 0.1 M, and 75 ml is 75 x 10-3 l, or 0.075 l.
∴ W = 0.5625 g
(Another way to do this is to ask how many grams would be required to make 1 l of the 0.1 M solution. So, X g ÷ 75 = 0.1 M, hence X = 75 x 0.1 = 7.5 g in 1 l is 0.1 M. If you have 7.5 g in 1 l, you must have 7.5 ÷ 1000 g/ml = 7.5 x 10-3 g/ml. You need 75 ml, so 75 x 7.5 x 10-3 g/ml = 0.5625 g.)
If you are struggling with chemistry calculations you may wish to take a look at chemCal, a basic chemistry calculator for the iPhone and iPod Touch.
A video demo of chemCal in operation is on YouTube, and the program is available on the iTunes App Store.
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How to make 25M water?
Well Anil, water has a molecular weight of 18 and a density of 1 g/ml. So, 1 liter is 1,000 g. The molarity of water is therefore 1000 / 18 = 55.6 M. The real question is: what do you want to use to dilute the water?