Question

Five different substances are given to you to be dissolved in water. which substances are most likely to undergo dissolution in water? check all that apply. view available hint(s) check all that apply. ethanol, ch3ch2oh potassium fluoride, kf octane, c8h18 toluene, c7h8 sodium iodide, nai

Answer 1

$\boxed{{\text{Ethanol, potassium fluoride and sodium iodide}}}$ are most likely to undergo dissolution in water.

Further explanation:

Solubility is the property of substance as a result of which it has a tendency to dissolve in other substances. It is measured in terms of the maximum amount of solute that can be dissolved in the given amount of solvent.

“Like dissolves like” is a general rule that is used to predict whether the substance is soluble in the given solvent or not.

Water is composed of one oxygen and two hydrogen atoms. Since O is more electronegative than H, water has some net dipole moment that is directed towards oxygen atom. So water is a polar molecule.

Ethanol $\left( {{\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_2}{\text{OH}}} \right)$ has a -OH group whose oxygen forms hydrogen bonding with the hydrogen atom of the water molecule and therefore ethanol is dissolved in water.

Potassium fluoride is an ionic compound that is composed of ${{\text{K}}^ + }$ and ${{\text{F}}^ - }$ ions. Water is a polar molecule that has ${{\text{H}}^ + }$ and ${{\text{F}}^ - }$ ions in it. So   is surrounded by ${\text{O}}{{\text{H}}^ - }$ and ${{\text{F}}^ - }$ is surrounded by ${{\text{H}}^ + }$ ions and therefore KF is dissolved in water.

Octane $\left( {{{\text{C}}_{\text{8}}}{{\text{H}}_{{\text{18}}}}} \right)$ is a hydrocarbon of eight carbon atoms and is non-polar in nature. So it does not dissolve in water.

Toluene $\left( {{{\text{C}}_{\text{7}}}{{\text{H}}_{\text{8}}}} \right)$ is an aromatic hydrocarbon that is non-polar in nature. So it does not dissolve in water.

Sodium iodide is an ionic compound that is composed of   and   ions. Water is a polar molecule that has ${{\text{H}}^ + }$ and ${\text{O}}{{\text{H}}^ - }$ ions in it. So ${\text{N}}{{\text{a}}^ + }$ is surrounded by ${\text{O}}{{\text{H}}^ - }$ and ${{\text{I}}^ - }$ is surrounded by ${{\text{H}}^ + }$ ions and therefore NaI is dissolved in water.

Therefore ethanol, potassium fluoride and sodium iodide are dissolved in water.

Learn more:

1. Identification of ionic bonding: brainly.com/question/1603987
2. What type of bond exists between phosphorus and chlorine? brainly.com/question/81715

Answer details:

Grade: High School

Chapter: Ionic and covalent compounds

Subject: Chemistry

Keywords: ethanol, potassium fluoride, sodium iodide, toluene, octane, water, polar, non-polar, hydrocarbon, dipole moment, NaI, KF, CH3CH2OH, C8H18, C7H8.

Answer 2

Correct Answer: Potassium Fluoride (KF) and sodium iodide (NaI)

Reason: 
Ionic compounds tends to undergo dissolution in aqueous medium. This results generation of ions. On other hand, covalent compounds donot generate ions in aqueous medium.

In present case, Potassium Fluoride (KF) and sodium iodide (NaI) are ionic compounds. Hence, when dissolved in water, these substances are most likely to undergo dissolution.