Solutions: Solubility and Intermolecular Forces Example 2

Jules Bruno
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identify the inter molecular forces present in both the solvent and solvent and predict whether a solution will form between the two. So here we have 50 g of arsenic Penta chloride placed into 250 g of water. Alright, So first of all, we know that water is our solvent since its large an amount warmer. We know as a polar solvent because it possesses hydrogen bonding. So we're gonna say h bonding now. Arsenic, Penta, chlorine. It's not as obvious what kind of inter Molecular forces has, so we're gonna draw it out. Arsenic is in group five A. So it has five valence electrons. And we're going to say here that the chlorine is in group 78, so they have seven valence electrons. And remember, halogen, when they're not in the center, only makes single bonds. So each one of these chlorine will single bond to the arsenic. And in that way, we'll have our molecular shape for arsenic Penta chloride. If we take a look here, we're gonna know that this is one of our perfect shapes that we've talked about in our earlier topics dealing with molecular polarity because it is a perfect shape. It has the same surrounding elements. It has no long pairs. It's one of the perfect shapes. It is non polar in nature. So we're going to say here that arsenic Penta chloride is non polar and therefore it's inter molecular force would be London Dispersion Forces or Van der Waals forces. Now London dispersion is a non polar force. Hydrogen bonding is a polar force. Their polarities don't match and as a result we cannot form a solution with them. So we've classified what their inter molecular forces are, and we've classified their polarity as a result. And from that we know that no A solution will not form between the two.