Doubts about iodine solution
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RT, to prepare an 0.1 mol/L iodine solution, we usually weigh out 13 g of iodine and dissolve it in potassium iodide, then dilute the mixture with water to a volume of 1 L. The molar mass of iodine is approximately 127, while the molecular weight of I2 is 254; therefore, 13/254 = 0.05. So the concentration of the resulting solution should be 0.05 mol/L. Then why is a solution prepared by using about 13 g of iodine in 1 L called 0.1 mol/L?
Calculatively, the amount of iodine (I2) is 0.05 mol, but when expressing its concentration we write it as C(1/2 I2) = 0.1 mol/L; whereas when expressed as C(I2), it is 0.05 mol/L. This is simply a difference in the ways concentrations are represented! This post was last edited by zjs0536 on 2009-3-23 21:48.]
Thanks for the advice, haha. I finally understand now
The concentration varies depending on the basic unit. For the same potassium permanganate solution, if the concentration is denoted as C, then it is equivalent to 5C when expressed in terms of 1/5 KMnO4.
One is molar concentration, and the other is equivalent concentration