Discuss The Effect Of Intermolecular Forces On Rate Of Solubility

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The effect of intermolecular forces on rate of solubility Introduction Solubility is a chemical property referring to the ability for a given substance, the solute, to dissolve in a solvent. When learning of this in class, the basics of this topic was explored, yet there was no deep exploration of it which intrigued me. For this IA, I am interested in delving into the factors which affect solubility, in this case I want to see the relationship between dipole moments and solubility. Each compound has intermolecular forces, the interaction between molecules in a compound. There are three types: London forces, dipole-dipole and hydrogen bonding. London forces are present in all compounds and are considered to be the weaker forces in comparison …show more content…

Meaning that the more polar a molecule is in its structure then it will have a greater dipole moment (Blaber 2017). From the solvents to be used it can be deduced that hexane must have the lowest dipole moment as it only has London forces meaning that it is not a polar substance. The dipole moment of the substance can be calculated by using the equation: μ=qr Where: μ is the dipole moment q is the amount of charge r is the distance between the charges (Ernest 2015) Clearly showing how the more polar a substance is, the greater its dipole moment will be, therefore in this experiment despite using the dipole moment as the independent variable, since the intermolecular forces are linked to the polarity of a molecule, then it will be researched whether a molecule with hydrogen bonds is a better solvent than a molecule with only London forces. However, for this experiment, the dipole moment will be found from an online source. From this, a graph can be plotted between dipole moment and the rate of …show more content…

To do so, the solute used in this experiment is NaCl. Amount of solute: since this experiment relies on the change in mass of the solute, to make this experiment more accurate, then the same amount of solute must be added to each solvent so it is easier to see which solvent dissolves the solute quicker. In this experiment 2 grams of NaCl will be added. Time of dissolving: to make the calculation of rate of solubility accurate, then the time given for the solvents to dissolve NaCl must be equal, in this experiment the time given will be 5 minutes. Amount of solvent: if there is a difference in the amount of solvent used for each trial, then the data collected will not be accurate due to the amount of solvent affecting rate of solubility. Therefore, to make this experiment fair, an equal amount of volume of the solvents will be used, which is 100

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