Comment on the solubility of iodine (I2) in each of the following liquids. State whether at room temperature you expect I2 to be highly soluble or almost insoluble, and choose the appropriate reasons why.
HINT: It may behoove you to draw the Lewis structure for each molecule.
(A) I2 in CCl4(liquid):
- unanswered
highly soluble almost insoluble
Explain your answer from part (A) above. Check all that apply.
- unanswered
The components are both polar and therefore have the capability of mixing. The components are both polar and therefore they don't have the capability of mixing. The components are both nonpolar and therefore have the capability of mixing. The components are both nonpolar and therefore they don't have the capability of mixing. One component is polar and the other is nonpolar, and therefore they have the capability of mixing. One component is polar and the other is nonpolar, and therefore they don't have the capability of mixing. The components both have similar molecular weights and therefore have the capability of mixing. The components both have similar molecular weights and therefore they don't have the capability of mixing. One component has a much higher molecular weight than the other, and therefore they have the capability of mixing. One component has a much higher molecular weight than the other, and therefore they don't have the capability of mixing. The components are both liquid at room temperature and therefore have the capability of mixing. The components are both liquid at room temperature and therefore they don't have the capability of mixing. The components both lack hydrogen bonding, and therefore have the capability of mixing. The components both lack hydrogen bonding, and therefore they don't have the capability of mixing. One component has hydrogen bonding, while the other lacks it; therefore, they have the capability of mixing. One component has hydrogen bonding, while the other lacks it; therefore, they don't have the capability of mixing.
(B) I2 in HF(liquid):
- unanswered
highly soluble almost insoluble
Explain your answer from part (B) above. Check all that apply.
- unanswered
The components are both polar and therefore have the capability of mixing. The components are both polar and therefore they don't have the capability of mixing. The components are both nonpolar and therefore have the capability of mixing. The components are both nonpolar and therefore they don't have the capability of mixing. One component is polar and the other is nonpolar, and therefore they have the capability of mixing. One component is polar and the other is nonpolar, and therefore they don't have the capability of mixing. The components both have similar molecular weights and therefore have the capability of mixing. The components both have similar molecular weights and therefore they don't have the capability of mixing. One component has a much higher molecular weight than the other, and therefore they have the capability of mixing. One component has a much higher molecular weight than the other, and therefore they don't have the capability of mixing. The components are both liquid at room temperature and therefore have the capability of mixing. The components are both liquid at room temperature and therefore they don't have the capability of mixing. The components both lack hydrogen bonding, and therefore have the capability of mixing. The components both lack hydrogen bonding, and therefore they don't have the capability of mixing. One component has hydrogen bonding, while the other lacks it; therefore, they have the capability of mixing. One component has hydrogen bonding, while the other lacks it; therefore, they don't have the capability of mixing.
A. highly solbule.
I2 dissolves easily in CCl4. Both are covalent compounds.
"The components are both nonpolar and therefore have the capability of mixing."
B.
Almost insoluble.
"One component is polar and the other is nonpolar, and therefore they don't have the capability of mixing."
Thank you so much for this!
sorry the text garbage..
a,b,c were correct while
"One component is polar and the other is nonpolar, and therefore they don't have the capability of mixing."
just thought you wanted to know, thank you again
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To determine the solubility of iodine (I2) in each of the liquids, we need to consider the polarity and molecular interactions between the components.
(A) I2 in CCl4 (liquid):
- I2 is a nonpolar molecule since the two iodine atoms are identical and share an equal distribution of electrons.
- CCl4 is also a nonpolar molecule because the carbon atom is surrounded by four chlorine atoms, creating a symmetrical arrangement that results in an equal distribution of electrons.
- Nonpolar molecules generally have weak intermolecular forces and are relatively insoluble in polar solvents.
Therefore, at room temperature, I2 is expected to be almost insoluble in CCl4 due to the nonpolar nature of both molecules.
(B) I2 in HF (liquid):
- I2 is still a nonpolar molecule, as mentioned above.
- HF is a polar molecule because the hydrogen atom bonded to fluorine creates a partial positive charge (δ+) on the hydrogen atom and a partial negative charge (δ-) on the fluorine atom. The difference in electronegativity between hydrogen and fluorine creates a polarity in the molecule.
Polar molecules tend to have stronger intermolecular forces and are more likely to be soluble in polar solvents.
Therefore, at room temperature, I2 is expected to be highly soluble in HF due to the polar nature of HF molecules.