You will need
- - table of solubility of substances.
The solution of this problem depends on the specific conditions. If you know the exact material mass and the exact volume of the solution, the solution is very simple. For example, 15 grams of barium chloride contains 400 milliliters of solution. What is its molar concentration?
Start with what you remember the exact formula of the salt: BaCl2. On the periodic table, determine the atomic masses of the elements included in its composition. And, given the index 2 from chlorine, get molecular weight: 137 + 71 = 208. Therefore, the molar mass of barium chloride 208 g/mol.
And in terms of the tasks, in the solution contains 15 grams of this substance. How much is it in moles? Dividing 15 by 208, will receive: approximately 0,072 mole.
Now you need to take into account that the solution volume to 1 liter, and 0.4. 0,072 dividing by 0.4, we get the answer: 0,18. That is, you have approximately 0.18-molar solution of barium chloride.
Let's build the solution. Suppose you began to dissolve in 100 milliliters of water at room temperature already mentioned, familiar to you, table salt is sodium chloride. Added you in small portions, stirring constantly and waiting until complete dissolution. And then came the moment when a small portion is not dissolved until the end, despite the intense mixing. You need to determine what is the molar concentration of the resulting solution.
First of all, you need to find the table of solubility of substances. They are in the majority of chemical directories, you can also find this information on the Internet. You can easily determine that at room temperature the saturation limit (i.e., solubility limit) of sodium chloride is 31.6 grams/100 grams of water.
For this task, you dissolved salt in 100 milliliters of water, but its density is almost equal 1. So we conclude: in the resulting solution contains approximately 31.6 grams of sodium chloride. Small undissolved excess, as well as some volume change upon dissolution of the salt can be neglected, the error will be small.
Respectively, in 1 liter of solution would contain 10 times more salt – 316 grams. Given that the molar mass of sodium chloride, as mentioned in the beginning, is 58.5 g/mol, you will easily find the answer: 316/58,5 = a 5.4-molar solution.