Step-by-step Solution:

1. Identify the chemical principle: A compound will be soluble in aqueous Sodium Hydroxide (NaOH) if it is acidic enough to react with NaOH in an acid-base reaction. The product of this reaction is typically a sodium salt, which is soluble in water. NaOH is a strong base, so it will react with acids that are stronger than water (pKa ≈ 15.7). The most common acidic organic compounds are carboxylic acids and phenols.

2. Analyze each compound: We will examine each compound from the given list to determine its acidity and reactivity with NaOH.

   • Phenol ($C_6H_5OH$): Phenols are acidic due to the resonance stabilization of the phenoxide ion formed after deprotonation. The pKa of phenol is approximately 10.0. Since this is much lower than the pKa of water, phenol is acidic enough to react with NaOH. $C_6H_5OH + NaOH \rightarrow C_6H_5O^-Na^+ + H_2O$ Sodium phenoxide is a salt and is soluble in water. Thus, phenol is soluble in aqueous NaOH.

   • Aniline ($C_6H_5NH_2$): Aniline is an aromatic amine. Amines are basic due to the lone pair of electrons on the nitrogen atom. They do not react with bases like NaOH. Thus, aniline is insoluble.

   • Benzoic acid ($C_6H_5COOH$): This is a carboxylic acid. Carboxylic acids are significantly acidic (pKa of benzoic acid is ~4.2). They readily react with strong bases like NaOH to form a water-soluble salt (sodium benzoate). $C_6H_5COOH + NaOH \rightarrow C_6H_5COO^-Na^+ + H_2O$ Thus, benzoic acid is soluble in aqueous NaOH.

   • Benzene ($C_6H_6$): Benzene is a non-polar hydrocarbon. It has no acidic protons and will not react with NaOH. Thus, benzene is insoluble.

   • Toluene ($C_6H_5CH_3$): Toluene is an alkylbenzene, a non-polar hydrocarbon. It is not acidic and does not react with NaOH. Thus, toluene is insoluble.

   • Benzyl alcohol ($C_6H_5CH_2OH$): This is an alcohol, not a phenol, as the -OH group is attached to a saturated carbon atom ($sp^3$ hybridized). Alcohols are very weakly acidic, with a pKa similar to water (~16-18). They are not acidic enough to react with NaOH to any significant extent. Thus, benzyl alcohol is insoluble.

   • Benzaldehyde ($C_6H_5CHO$): This is an aldehyde. It does not have any acidic protons that can be removed by NaOH. Although it can undergo the Cannizzaro reaction in concentrated NaOH, it does not dissolve through a simple acid-base reaction. Thus, benzaldehyde is insoluble.

   • p-Nitrophenol: This is a substituted phenol. The nitro group ($-NO_2$) is a strong electron-withdrawing group. It stabilizes the phenoxide ion through resonance, making p-nitrophenol more acidic than phenol (pKa ≈ 7.1). It will readily react with NaOH. $O_2N-C_6H_4-OH + NaOH \rightarrow O_2N-C_6H_4-O^-Na^+ + H_2O$ Thus, p-nitrophenol is soluble in aqueous NaOH.

   • m-Cresol (3-methylphenol): This is also a substituted phenol. The methyl group ($-CH_3$) is a weak electron-donating group, which makes m-cresol slightly less acidic than phenol (pKa ≈ 10.1). However, it is still a phenol and is acidic enough to react with NaOH. $CH_3-C_6H_4-OH + NaOH \rightarrow CH_3-C_6H_4-O^-Na^+ + H_2O$ Thus, m-cresol is soluble in aqueous NaOH.

3. Count the soluble compounds: The compounds that are soluble in aqueous NaOH are:

   1. Phenol
   2. Benzoic acid
   3. p-Nitrophenol
   4. m-Cresol

   The total number of soluble compounds is 4.

Final Answer

The total number of compounds soluble in aqueous NaOH is 4.