Verified Answer :
There are solubility rules for inorganic compounds in water as shown in the picture attached. For organic compounds, we use the concept of polarity.
1. NaCl: Since NaCl is a chloride and Na is not included in the list of insoluble, therefore, NaCl is highly soluble.
2. CH₃Cl: This compound is organic. Since the electonegativity between C and Cl is great, it tends to be polar. Because water is also polar, this is highly soluble.
3. CH₃OH: This is an alcohol. Alcohols with short chains of hydrocarbon are highly soluble in water because of hydrogen bonding (-OH).
4. KOH: From the table, KOH is highly soluble. In fact, this is a strong solid which readily dissociates in water.
5. C₆H₆ is the benzene ring. This is a nonpolar molecule, so you would expect this to be insoluble with water.
6. C₆H₁₄ is hexane which is a long chain of hydrocarbons. Because there is no point of hydrogen bonding, this is nonpolar, and therefore, insoluble in water.
7. KNO₃ is highly soluble because all nitrates are soluble.
8. CCl₄ is an organic compound. Individually, C-Cl bonds are polar. But because there are 4 of them, they cancel out. As a result, there is a balance of partial charges which makes it nonpolar. Thus, this is insoluble in water.
9. NaSO₄: from the table, this is highly soluble.
10. CH₃CH₂OH: Although it is composed of the nonpolar hydrocarbon chain, the OH group is much stronger such that it is highly soluble in water.
11. CH₂Cl₂: This is polar because there is an imbalance due to the C-H and C-Cl bonds. Thus, it is highly soluble in water.
12. CH₃COOH is a carboxylic acid which is even more soluble than alcohols. Thus, this is highly soluble in water.
NaCl, KOH, \(\text{KNO}_3\), \(\text{NA}_2\text{SO}_4\), \(\text{CH}_3\text{Cl}\), \(\text{CH}_2\text{Cl}_2\), \(\text{CH}_3\text{OH}\), \(\text{CH}_3\text{CH}_2\text{OH}\) and \(\text{CH}_3\text{COOH}\) are highly soluble in water whereas \(\text{C}_6\text{H}_6\), \({\text{C}_6\text{H}_14}\), and \({\text{CC}}{{\text{l}}_{\text{4}}}\) are highly insoluble in water.
Further explanation:
Solubility is the property of substance as a result of which it has a tendency to dissolve in other substances. “Like dissolves like” is a general principle that is used to predict whether the substance is soluble in the given solvent or not.
Water is a polar molecule and it has \(\text{H}^+\) and \(\text{OH}^-\) ions in it. So ionic and polar compounds are highly soluble in water whereas the covalent and organic compounds are highly insoluble in water.
NaCl, KOH, \({\text{KN}}{{\text{O}}_{\text{3}}}\) and \({\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\) are ionic compounds that are composed of their respective ions. Water is a polar compound and therefore all these compounds are highly soluble in water.
\({{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}\) and \({{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{14}}}}\) are non-polar hydrocarbons so these are highly insoluble in water.
\({\text{C}}{{\text{H}}_3}{\text{Cl}}\) and \({\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}\) are polar covalent compounds and water is also polar. So these are highly soluble in water.
\({\text{CC}}{{\text{l}}_{\text{4}}}\) is a covalent compound so it cannot form ions and therefore it is highly insoluble in water.
\({\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}\) and \({\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH}}\) are alcohols that contains a highly electronegative oxygen atom that is capable to form hydrogen bonds with the hydrogen atom of water molecule. Therefore both these compounds are highly soluble in water.
\({\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}\) is polar due to presence of hydroxyl group in it and therefore it is highly soluble in water.
Therefore, NaCl, KOH, \({\text{KN}}{{\text{O}}_{\text{3}}}\), \({\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\), \({\text{C}}{{\text{H}}_3}{\text{Cl}}\), \({\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}\), \({\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}\), \({\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH}}\) and \({\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}\) are highly soluble in water whereas \({{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}\) , \({{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{14}}}}\) , and \({\text{CC}}{{\text{l}}_{\text{4}}}\) are highly insoluble in water.
Learn more:
- Identification of ionic bonding: https://brainly.com/question/1603987
- What type of bond exists between phosphorus and chlorine? https://brainly.com/question/81715
Answer details:
Grade: High School
Chapter: Ionic and covalent compounds
Subject: Chemistry
Keywords: NaCl, KOH, KNO3, CH3OH, CH3CH2OH, CH3COOH, C6H6, C6H14, CCl4, CH3Cl, Na2SO4, CH2Cl2, soluble, insoluble, water.