Step 1: Identifying Given Variables

To determine the mass percentage of the solute in the solution, we define the following variables based on the provided data:

• Mass of solute (aspirin, $C_{9}H_{8}O_{4}$), $m_{solute} = 6.5\text{ g}$
• Mass of solvent (acetonitrile, $CH_{3}CN$), $m_{solvent} = 450\text{ g}$

Step 2: Calculating Total Mass of the Solution

The total mass of the solution is the sum of the masses of the solute and the solvent [Per the principle of additivity of mass in binary solutions]:

$m_{solution} = m_{solute} + m_{solvent}$

$m_{solution} = 6.5\text{ g} + 450\text{ g} = 456.5\text{ g}$

Step 3: Defining the Mass Percentage Formula

The mass percentage of a component in a solution is defined as the mass of the component per 100 grams of the solution [Definition of Mass/Mass Percent Concentration]:

$\text{Mass Percent} = \left( \frac{m_{solute}}{m_{solution}} \right) \times 100$

Step 4: Algebraic Substitution and Execution

Substitute the established values into the formula:

$\text{Mass Percent} = \left( \frac{6.5\text{ g}}{456.5\text{ g}} \right) \times 100$

$\text{Mass Percent} \approx 0.01423877 \times 100$

$\text{Mass Percent} \approx 1.423877\%$

Step 5: Significant Figures and Final Rounding

Given that the input values $6.5\text{ g}$ (2 significant figures) and $450\text{ g}$ (2 or 3 significant figures, assuming 2 based on the precision of the solute mass) are provided, the final result should be rounded to two significant figures to maintain scientific rigor [Per standard laboratory reporting protocols].

$1.423877\% \approx 1.4\%$

Final Solution: The mass percentage of aspirin in the acetonitrile solution is 1.42%.

1. Definition of Solution

A solution is defined as a homogeneous mixture of two or more chemically non-reacting substances whose composition can be varied within certain limits. The component present in the larger quantity is termed the solvent, while the component(s) present in smaller quantities are termed solutes. [Per the fundamental principles of chemical thermodynamics and phase behavior].

2. Classification of Solutions

Solutions are classified based on the physical state of the solvent and the solute. Since there are three states of matter (solid, liquid, gas), there are nine possible types of solutions. The following table delineates these categories:

3. Brief Analysis of Solution Types

• Gaseous Solutions: The solvent is always a gas. These mixtures are generally ideal or near-ideal if the gases do not interact chemically. [Refer to Dalton's Law of Partial Pressures for gas mixture dynamics].
• Liquid Solutions: The solvent is a liquid. These are the most common solutions in chemistry. The solute can be a gas (e.g., carbonated beverages), a liquid (e.g., alcoholic drinks), or a solid (e.g., saline solutions).
• Solid Solutions: The solvent is a solid. These involve either interstitial solid solutions (small atoms occupying gaps in the crystal lattice) or substitutional solid solutions (atoms replacing solvent atoms in the lattice).

Final Solution: A solution is a homogeneous mixture of two or more components. There are 9 distinct types of solutions based on the physical states (solid, liquid, gas) of the solute and solvent, categorized as gaseous, liquid, or solid solutions respectively.

Step 1: Chemical Structure and Composition Analysis

Toluene, represented by the chemical formula $C_6H_5CH_3$, consists of a hydrophobic phenyl ring ($C_6H_5-$) attached to a methyl group ($-CH_3$). [Per the principles of molecular structure, this constitutes a non-polar hydrocarbon derivative.]

Step 2: Evaluating Solubility in Water ($H_2O$)

The solubility of a substance in water is governed by the principle of "like dissolves like." Water is a highly polar solvent characterized by an extensive intermolecular hydrogen-bonding network. To dissolve a substance, the solute must be able to:

• Form hydrogen bonds with water molecules, or
• Possess a significant dipole moment that overcomes the energy required to disrupt the hydrogen bonds between $H_2O$ molecules.

Step 3: Analyzing Intermolecular Forces

Toluene possesses only weak London dispersion forces and very weak dipole-dipole interactions due to the minor electron-donating effect of the methyl group. It is essentially non-polar. When toluene is introduced to water, the energy released upon the formation of weak solute-solvent interactions is significantly less than the energy required to break the strong hydrogen bonds between water molecules. [Per the Thermodynamic Criterion for Solubility: $\Delta G_{sol} = \Delta H_{sol} - T\Delta S_{sol}$, where the positive enthalpy change ($\Delta H > 0$) is not compensated by sufficient entropy change to make the dissolution process spontaneous.]

Step 4: Visualizing the Interaction (Molecular Exclusion)

Conclusion

Because toluene lacks polar functional groups (like $-OH$ or $-NH_2$) that can participate in hydrogen bonding with water, it remains immiscible. Hydrocarbon chains of this type are classified as hydrophobic.

Final Solution: Toluene is insoluble in water.

Step 1: Theoretical Foundation of Freezing Point Depression

The depression in freezing point ($\Delta T_f$) is a colligative property defined by the expression:

$\Delta T_f = i \cdot K_f \cdot m$

Where:

• $i$ is the van't Hoff factor.
• $K_f$ is the cryoscopic constant of the solvent (water).
• $m$ is the molality of the solution.

[Per Raoult’s Law and the theory of colligative properties, $\Delta T_f$ is directly proportional to the total number of solute particles present in the solution.]

Step 2: Analysis of Acid Dissociation

Acetic acid ($CH_3COOH$) and its halogenated derivatives ($CCl_3COOH$ and $CF_3COOH$) act as weak acids in water. They undergo partial dissociation represented by the equilibrium:

$R-COOH \rightleftharpoons R-COO^- + H^+$

The extent of dissociation determines the value of $i$. Since $i = 1 + \alpha(n-1)$, where $\alpha$ is the degree of dissociation and $n=2$, a higher degree of dissociation ($\alpha$) results in a larger van't Hoff factor ($i$).

Step 3: Evaluating Inductive Effects (-I Effect)

The acidity of carboxylic acids is dictated by the stability of the conjugate base ($R-COO^-$). The strength of the acid increases with the electron-withdrawing nature of the substituent group attached to the carboxyl group.

[By the Electronic Displacement Theory, the $-I$ effect stabilizes the carboxylate anion by dispersing negative charge. Electronegativity of Fluorine ($3.98$) > Chlorine ($3.16$), therefore, $-CF_3$ is more electron-withdrawing than $-CCl_3$.]

Step 4: Comparative Ionization and Resulting $\Delta T_f$

The acidity order is: $CH_3COOH < CCl_3COOH < CF_3COOH$.

1. $CH_3COOH$: Weakest acid, lowest degree of dissociation ($\alpha$), smallest $i$.
2. $CCl_3COOH$: Stronger acid due to the $-I$ effect of three chlorine atoms, higher $\alpha$, larger $i$.
3. $CF_3COOH$: Strongest acid due to the intense $-I$ effect of three fluorine atoms, highest $\alpha$, largest $i$.

Since $\Delta T_f \propto i$, the order of depression in freezing point must follow the order of the van't Hoff factor: $CH_3COOH < CCl_3COOH < CF_3COOH$.

Final Solution: The increase in depression of freezing point is attributed to the increased strength of the acids due to the strong electron-withdrawing inductive effect of the halogen atoms, which leads to a greater degree of dissociation ($i$) in the sequence $CH_3COOH < CCl_3COOH < CF_3COOH$.

Step 1: Theoretical Foundation of Solubility

The solubility product constant ($K_{sp}$) describes the equilibrium between a sparingly soluble ionic compound and its saturated solution. For the dissociation of Copper(II) sulfide (CuS) in water, the chemical equilibrium is expressed as:

$CuS(s) \rightleftharpoons Cu^{2+}(aq) + S^{2-}(aq)$

[Per the Law of Mass Action, the $K_{sp}$ expression is the product of the molar concentrations of the constituent ions raised to their stoichiometric coefficients].

Step 2: Defining Molar Solubility

Let $s$ represent the molar solubility of CuS in $mol \cdot L^{-1}$. When $s$ moles of CuS dissolve, they produce $s$ moles of $Cu^{2+}$ ions and $s$ moles of $S^{2-}$ ions per liter of solution.

• $[Cu^{2+}] = s$
• $[S^{2-}] = s$

Step 3: Setting up the Algebraic Equation

The expression for the solubility product is:

$K_{sp} = [Cu^{2+}][S^{2-}]$

Substituting the molar solubility values into the equation:

$K_{sp} = (s)(s) = s^2$

Step 4: Calculation of Molar Solubility

Given that $K_{sp} = 6 \times 10^{-16}$, we equate this to our derived expression:

$s^2 = 6 \times 10^{-16}$

To solve for $s$, we take the square root of both sides:

$s = \sqrt{6 \times 10^{-16}}$

$s = \sqrt{6} \times \sqrt{10^{-16}}$

$s \approx 2.449 \times 10^{-8} \, mol \cdot L^{-1}$

Step 5: Illustrative Equilibrium Diagram

Final Analytical Synthesis

Final Solution: The maximum molarity of CuS in an aqueous solution is $2.45 \times 10^{-8} \, mol \cdot L^{-1}$.

1. Theoretical Basis: Raoult's Law

According to Raoult's Law, for an ideal solution, the partial vapor pressure ($p_i$) of each volatile component in the solution is directly proportional to its mole fraction ($x_i$) in the solution: $p_1 = p_1^0 x_1$ and $p_2 = p_2^0 x_2$. Solutions that deviate from this ideal behavior are termed non-ideal solutions.

2. Positive Deviation from Raoult's Law

A solution exhibits positive deviation when the observed vapor pressure is higher than the vapor pressure calculated based on Raoult’s Law.

• Intermolecular Forces: The solute-solvent ($A-B$) interaction is weaker than the solute-solute ($A-A$) and solvent-solvent ($B-B$) interactions. [Per Molecular Kinetic Theory]
• Molecular Escape: Because $A-B$ forces are weaker, the molecules escape more easily into the vapor phase, increasing the total vapor pressure.
• Enthalpy of Mixing ($\Delta_{mix}H$): Since energy must be absorbed to overcome the stronger $A-A$ and $B-B$ interactions to form weaker $A-B$ interactions, the process is endothermic. Thus, $\Delta_{mix}H > 0$.
• Volume of Mixing ($\Delta_{mix}V$): The weak interaction leads to an expansion of the liquid, resulting in $\Delta_{mix}V > 0$.

3. Negative Deviation from Raoult's Law

A solution exhibits negative deviation when the observed vapor pressure is lower than the vapor pressure calculated based on Raoult’s Law.

• Intermolecular Forces: The solute-solvent ($A-B$) interaction is stronger than the solute-solute ($A-A$) and solvent-solvent ($B-B$) interactions.
• Molecular Escape: The stronger $A-B$ attractive forces hold molecules within the liquid phase, decreasing the tendency to escape into the vapor phase.
• Enthalpy of Mixing ($\Delta_{mix}H$): Energy is released when stronger $A-B$ bonds are formed, making the process exothermic. Thus, $\Delta_{mix}H < 0$.
• Volume of Mixing ($\Delta_{mix}V$): The strong interaction leads to contraction of the liquid, resulting in $\Delta_{mix}V < 0$.

4. Comparative Summary Table

5. Graphical Representation of Deviations

Final Solution: Positive deviation occurs when intermolecular forces are weakened ($\Delta_{mix}H > 0$), while negative deviation occurs when intermolecular forces are strengthened ($\Delta_{mix}H < 0$).

Step 1: Establishing the Theoretical Framework

To determine the vapor pressure of pure water ($P_1^\circ$), we utilize Raoult's Law for solutions containing a non-volatile solute. The relative lowering of vapor pressure is defined by the mole fraction of the solute ($x_2$):

$\frac{P_1^\circ - P_1}{P_1^\circ} = x_2 = \frac{n_2}{n_1 + n_2}$ [Per Raoult's Law]

Where:

• $P_1^\circ$: Vapor pressure of pure water.
• $P_1$: Vapor pressure of the solution.
• $n_1$: Moles of solvent ($H_2O$, Molar mass = $18 \text{ g/mol}$).
• $n_2$: Moles of solute (Molar mass = $M_2$).

Step 2: Formulating System Equations

Initial State (Solution A):

Mass of solute ($w_2$) = $30 \text{ g}$

Mass of solvent ($w_1$) = $90 \text{ g}$

$n_1 = \frac{90}{18} = 5 \text{ mol}$

$n_2 = \frac{30}{M_2}$

$\frac{P_1^\circ - 2.8}{P_1^\circ} = \frac{30/M_2}{5 + 30/M_2}$ --- (Equation 1)

Final State (Solution B after adding 18 g water):

New mass of solvent ($w_1'$) = $90 + 18 = 108 \text{ g}$

$n_1' = \frac{108}{18} = 6 \text{ mol}$

$\frac{P_1^\circ - 2.9}{P_1^\circ} = \frac{30/M_2}{6 + 30/M_2}$ --- (Equation 2)

Step 3: Algebraic Simplification

Let $x = \frac{30}{M_2}$. Rewrite Equations 1 and 2:

Eq 1: $1 - \frac{2.8}{P_1^\circ} = \frac{x}{5+x} \implies \frac{2.8}{P_1^\circ} = 1 - \frac{x}{5+x} = \frac{5}{5+x} \implies 5+x = \frac{5P_1^\circ}{2.8}$

Eq 2: $1 - \frac{2.9}{P_1^\circ} = \frac{x}{6+x} \implies \frac{2.9}{P_1^\circ} = 1 - \frac{x}{6+x} = \frac{6}{6+x} \implies 6+x = \frac{6P_1^\circ}{2.9}$

Step 4: Solving for $P_1^\circ$

Subtracting the expression for $(5+x)$ from $(6+x)$:

$(6 + x) - (5 + x) = \frac{6P_1^\circ}{2.9} - \frac{5P_1^\circ}{2.8}$

$1 = P_1^\circ \left( \frac{6}{2.9} - \frac{5}{2.8} \right)$

$1 = P_1^\circ \left( \frac{16.8 - 14.5}{8.12} \right)$

$1 = P_1^\circ \left( \frac{2.3}{8.12} \right)$

$P_1^\circ = \frac{8.12}{2.3} \approx 3.53 \text{ kPa}$

Data Visualization of Vapor Pressure Relationship

Final Solution: The vapor pressure of water at 298 K is $3.53 \text{ kPa}$.

Definition of Mass Percentage

Mass percentage ($w/w\%$) is a concentration unit defined as the mass of a specific solute component in grams present per $100$ grams of the total solution. It represents the relative mass of a component relative to the total mass of the mixture. [Per IUPAC Compendium of Chemical Terminology]

Mathematical Formulation

For a solution containing a solute $A$ and a solvent $B$, the mass percentage of component $A$ is calculated using the following expression:

$\text{Mass percentage of component } A = \left( \frac{\text{Mass of component } A}{\text{Total mass of the solution}} \right) \times 100$

Variable Identification

• $W_A$: Mass of solute component $A$ (in grams).
• $W_B$: Mass of solvent component $B$ (in grams).
• $W_{\text{total}}$: Total mass of the solution ($W_A + W_B$).

Procedural Calculation

To determine the mass percentage, follow these granular steps:

1. Summation of Masses: Calculate the total mass of the solution by aggregating the masses of all individual components: $W_{\text{total}} = W_A + W_B$. [By the Law of Conservation of Mass]
2. Fractional Determination: Divide the mass of the solute by the total mass of the solution to obtain the mass fraction: $X_A = \frac{W_A}{W_A + W_B}$.
3. Percentage Conversion: Multiply the obtained mass fraction by $100$ to express the value as a percentage.

Application Context

Mass percentage is particularly useful in industrial chemistry and commercial products (such as bleach or household reagents) because it is independent of temperature variations, unlike molarity or normality which rely on volume. [Per the definition of concentration units in chemical thermodynamics]

Final Solution: The mass percentage of a component is defined as the mass of that component expressed as a percentage of the total mass of the solution, mathematically denoted by $\frac{W_{\text{solute}}}{W_{\text{solution}}} \times 100$.

Step 1: Identifying the Governing Law

The solubility of a gas in a liquid at a constant temperature is directly proportional to the partial pressure of the gas above the liquid. This relationship is defined by Henry’s Law:

$p = K_H \cdot x$

Where $p$ is the partial pressure, $K_H$ is Henry's Law constant, and $x$ is the mole fraction of the gas in the solution. Since the amount of solvent is constant in both cases, the mole fraction $x$ is directly proportional to the mass of the solute (ethane). Therefore, we can simplify the relationship to:

$p \propto m$ or $\frac{p_1}{m_1} = \frac{p_2}{m_2}$ [Per the linear proportionality defined by Henry's Law at constant temperature]

Step 2: Given Parameters

Step 3: Algebraic Calculation

Using the proportionality derived in Step 1:

$\frac{p_1}{m_1} = \frac{p_2}{m_2}$

Rearranging to solve for $p_2$:

$p_2 = \frac{p_1 \cdot m_2}{m_1}$

Substituting the values:

$p_2 = \frac{1 \text{ bar} \cdot (5.00 \times 10^{-2} \text{ g})}{6.56 \times 10^{-3} \text{ g}}$

Step 4: Arithmetic Simplification

Perform the division of the mass values:

$p_2 = \frac{5.00 \times 10^{-2}}{6.56 \times 10^{-3}}$

$p_2 = \frac{50.0 \times 10^{-3}}{6.56 \times 10^{-3}}$ [By shifting decimal point to align exponents]

$p_2 = 7.6219... \text{ bar}$

Visual Representation of Solubility Proportionality

Final Solution: The partial pressure of the ethane gas will be 7.62 bar.

1. Formal Definition of Molarity

Molarity ($M$) is defined as the number of moles of a solute dissolved per liter of the total solution. It is a temperature-dependent unit of concentration, as the volume of the solution fluctuates with variations in temperature [Per the principle of thermal expansion].

2. Mathematical Formulation

The molarity of a solution is represented by the following mathematical expression:

$M = \frac{n_{solute}}{V_{solution(L)}}$

Where:

• $n_{solute}$ = Number of moles of the solute.
• $V_{solution(L)}$ = Volume of the total solution measured in liters ($L$).

3. Derivation from Mass and Molar Mass

Given that the number of moles ($n$) is calculated by the ratio of the given mass of the substance ($w$) to its molar mass ($M_m$), we substitute this into the primary equation:

$M = \frac{w_B}{M_{m(B)} \times V_{solution(L)}}$

If the volume is provided in milliliters ($mL$), the conversion factor is applied to normalize the units to liters:

$M = \frac{w_B \times 1000}{M_{m(B)} \times V_{solution(mL)}}$

4. Conceptual Visualization

The following diagram represents the transition from a pure solute to a solution of defined molarity.

5. Dimensional Analysis

Final Solution: Molarity is the concentration unit representing the moles of solute per liter of solution, denoted as $M = \frac{n}{V(L)}$. It is highly sensitive to temperature changes due to the liquid expansion coefficient of the solvent.

Step 1: Theoretical Foundation

According to Raoult's Law for a solution containing a non-volatile solute, the relative lowering of vapour pressure is equal to the mole fraction of the solute in the solution. The mathematical expression is defined as:

$\frac{P_1^\circ - P_1}{P_1^\circ} = x_2 = \frac{n_2}{n_1 + n_2}$

Where:

• $P_1^\circ$ = Vapour pressure of pure solvent
• $P_1$ = Vapour pressure of solution
• $n_2$ = Moles of solute
• $n_1$ = Moles of solvent (octane)

Step 2: Defining Given Parameters

Step 3: Calculating Mole Fraction Constraints

Given the condition $\frac{P_1}{P_1^\circ} = 0.8$, the relative lowering of vapour pressure is:

$\frac{P_1^\circ - 0.8 P_1^\circ}{P_1^\circ} = \frac{0.2 P_1^\circ}{P_1^\circ} = 0.2$

Thus, $x_2 = 0.2$. [By Raoult's Law definition of relative lowering of vapour pressure].

Step 4: Algebraic Derivation for Solute Mass

The mole fraction of solute $x_2$ is expressed as:

$x_2 = \frac{n_2}{n_1 + n_2} = 0.2$

First, calculate moles of solvent ($n_1$):

$n_1 = \frac{W_1}{M_1} = \frac{114 \, \text{g}}{114 \, \text{g mol}^{-1}} = 1 \, \text{mol}$

Substitute $n_1$ into the mole fraction equation:

$\frac{n_2}{1 + n_2} = 0.2$

$n_2 = 0.2(1 + n_2)$

$n_2 = 0.2 + 0.2 n_2$

$0.8 n_2 = 0.2$

$n_2 = \frac{0.2}{0.8} = 0.25 \, \text{mol}$

Step 5: Final Mass Calculation

Using the relation $n_2 = \frac{W_2}{M_2}$:

$W_2 = n_2 \times M_2$

$W_2 = 0.25 \, \text{mol} \times 40 \, \text{g mol}^{-1}$

$W_2 = 10 \, \text{g}$

Final Solution: The mass of the non-volatile solute required is 10 g.

Definition of Mass Percentage ($w/w\%$)

Mass percentage is a concentration unit used to express the composition of a mixture. It defines the mass of a specific component present in $100$ grams of the total solution. This unit is temperature-independent because mass does not fluctuate with thermal changes [Per the Law of Conservation of Mass].

Mathematical Formulation

For a binary solution containing a solute $A$ and a solvent $B$, the mass percentage of the solute is defined as the ratio of the mass of the solute to the total mass of the solution, multiplied by $100$.

The governing equation is:

$\text{Mass Percentage of component} = \left( \frac{\text{Mass of component in the solution}}{\text{Total mass of the solution}} \right) \times 100$

Symbolically:

$w_A \% = \frac{m_A}{m_A + m_B} \times 100$

• $m_A$: Mass of the solute in grams ($g$).
• $m_B$: Mass of the solvent in grams ($g$).
• $m_A + m_B$: Total mass of the solution in grams ($g$).

Visual Representation of Mass Composition

Comparative Analysis of Concentration Units

Conceptual Application

If a solution is labeled as $10\% \text{ glucose in water by mass}$, it signifies that $10\text{ g}$ of glucose is dissolved in $90\text{ g}$ of water, resulting in $100\text{ g}$ of total solution [By additive property of mass].

Final Solution: Mass percentage is the mass of a component expressed as a fraction of the total mass of the mixture, multiplied by 100, represented as $\text{Mass \%} = \frac{m_{\text{component}}}{m_{\text{solution}}} \times 100$.

1. Chemical Structure and Molecular Analysis of Phenol ($C_6H_5OH$)

Phenol consists of a polar hydroxyl ($-OH$) group attached to a non-polar hydrophobic phenyl ring ($C_6H_5-$). The solubility of an organic compound in water is determined by the competition between the tendency of the polar group to form hydrogen bonds with water molecules and the tendency of the non-polar hydrocarbon part to disrupt the hydrogen-bonded structure of water.

2. Interaction Mechanism with Solvent (Water)

• Hydrophilic interaction: The $-OH$ group in phenol is capable of forming intermolecular hydrogen bonds with water molecules ($R-OH \cdots OH_2$). [Per the principles of Hydrogen Bonding in polar solvents].
• Hydrophobic interaction: The aromatic phenyl ring is bulky and lipophilic. It cannot form hydrogen bonds with water and effectively creates a "solvent cage" that decreases the entropy of the water system, which is energetically unfavorable. [Per the Thermodynamic principle of hydrophobic effect].

3. Quantitative Solubility Assessment

At room temperature ($25^\circ C$ or $298 K$), the solubility of phenol in water is approximately $8.3 \, g$ per $100 \, mL$ of water. This specific value allows for a formal classification within the solubility continuum:

4. Graphical Representation of Solubility Dynamics

5. Justification of Conclusion

Because the hydrophobic nature of the benzene ring is significant enough to counteract the polar $-OH$ group, but not strong enough to render the compound entirely hydrophobic, phenol occupies the intermediate classification. It exhibits limited miscibility, separating into two phases if the concentration exceeds the solubility limit of $8.3 \, g/100 \, mL$. [Per the 'Like Dissolves Like' principle applied to amphiphilic molecules].

Final Solution: Phenol is partially soluble in water.

Step 1: Chemical Formula and Molar Mass Calculation

The solute is 2-chlorobutanoic acid, $\text{CH}_{3}\text{CH}_{2}\text{CHClCOOH}$. The chemical formula is $\text{C}_{4}\text{H}_{7}\text{ClO}_{2}$.

Calculation of Molar Mass ($M_2$):

• $4 \times \text{C} = 4 \times 12.01 = 48.04 \text{ g/mol}$
• $7 \times \text{H} = 7 \times 1.008 = 7.056 \text{ g/mol}$
• $1 \times \text{Cl} = 35.45 \text{ g/mol}$
• $2 \times \text{O} = 2 \times 16.00 = 32.00 \text{ g/mol}$
• Total $M_2 = 122.55 \text{ g/mol}$

Step 2: Molality ($m$) Calculation

Molality is defined as the moles of solute per kilogram of solvent [Per the definition of molality].

Mass of solute ($w_2$) = $10 \text{ g}$

Mass of solvent ($w_1$) = $250 \text{ g} = 0.250 \text{ kg}$

$m = \frac{w_2}{M_2 \times w_1 (\text{kg})} = \frac{10}{122.55 \times 0.250} \approx 0.3264 \text{ mol/kg}$

Step 3: Determination of Van't Hoff Factor ($i$)

Since $\text{CH}_{3}\text{CH}_{2}\text{CHClCOOH}$ is a weak acid, it undergoes partial dissociation:

$\text{CH}_{3}\text{CH}_{2}\text{CHClCOOH} \rightleftharpoons \text{CH}_{3}\text{CH}_{2}\text{CHClCOO}^- + \text{H}^+$

Degree of dissociation ($\alpha$) is calculated using Ostwald's Dilution Law for weak electrolytes:

$\alpha = \sqrt{\frac{K_a}{C}}$

Given $K_a = 1.4 \times 10^{-3}$ and $C \approx m = 0.3264 \text{ M}$:

$\alpha = \sqrt{\frac{1.4 \times 10^{-3}}{0.3264}} = \sqrt{0.004289} \approx 0.0655$

The Van't Hoff factor $i = 1 + \alpha$ [Per the relation $i = 1 + (n-1)\alpha$, where $n=2$]:

$i = 1 + 0.0655 = 1.0655$

Step 4: Calculation of Depression in Freezing Point ($\Delta T_f$)

Applying the Colligative Property formula [Per Raoult’s Law and Van't Hoff modifications]:

$\Delta T_f = i \cdot K_f \cdot m$

$\Delta T_f = 1.0655 \times 1.86 \text{ K kg mol}^{-1} \times 0.3264 \text{ mol kg}^{-1}$

$\Delta T_f = 0.6467 \text{ K}$

Visual Representation of Dissociation Equilibrium

Final Solution: The depression in the freezing point of water is 0.647 K.

Definition of Henry's Law

Henry’s Law states that at a constant temperature, the partial pressure of a gas ($p$) in vapor phase is directly proportional to the mole fraction of the gas ($x$) in the solution. This is mathematically expressed as:

$p = K_H \cdot x$

Where:

• $p$ = Partial pressure of the gas above the liquid.
• $x$ = Mole fraction of the gas in the solution.
• $K_H$ = Henry's Law constant (units of pressure, e.g., $\text{bar}$ or $\text{atm}$).

[Per the kinetic molecular theory, $K_H$ is a function of the nature of the gas and the solvent; as temperature increases, $K_H$ increases, leading to a decrease in the solubility of the gas].

Applications of Henry's Law

Final Solution: Henry's Law establishes that gas solubility in a liquid is proportional to its partial pressure ($p = K_H \cdot x$). Its critical applications include the carbonation of beverages, the mitigation of decompression sickness in deep-sea diving, and the physiological challenges faced by climbers at high altitudes due to low oxygen partial pressure.

Step 1: Analyzing the Solute and Solvent Characteristics

To identify the intermolecular interactions, we must first decompose the species into their fundamental ionic or molecular components:

• Solute: $NaClO_4$ (Sodium perchlorate). This is an ionic salt. Upon dissolution in a polar solvent, it dissociates completely into its constituent ions: $NaClO_{4(s)} \rightarrow Na^+_{(aq)} + ClO_{4}^-_{(aq)}$ [Based on the Arrhenius theory of electrolytic dissociation].
• Solvent: $H_2O$ (Water). Water is a polar protic solvent characterized by a high dielectric constant ($\varepsilon \approx 78.4$ at $298K$). It possesses a permanent dipole moment due to the electronegativity difference between Oxygen and Hydrogen.

Step 2: Identifying the Interaction Mechanism

When the ionic solid $NaClO_4$ is introduced to $H_2O$, the electrostatic forces holding the crystal lattice ($Na^+$ and $ClO_4^-$ ions) are overcome by the strong attractive forces exerted by the water molecules.

• The partial negative charge ($\delta^-$) on the Oxygen atom of $H_2O$ is attracted to the $Na^+$ cation.
• The partial positive charge ($\delta^+$) on the Hydrogen atoms of $H_2O$ is attracted to the $ClO_4^-$ anion.

This type of interaction, where an ion is surrounded and stabilized by solvent molecules, is defined as an Ion-Dipole Interaction [Per the electrostatic principles of Coulomb's Law, $F = k \frac{q_1 q_2}{r^2}$].

Step 3: Visualizing the Ion-Dipole Interaction

Step 4: Comparative Summary of Interactions

Final Solution: The most important type of intermolecular attractive interaction in the pair $NaClO_4$ and water is the Ion-Dipole interaction.

1. Definition of Molality ($m$)

Molality is defined as the number of moles of a solute present in $1\text{ kg}$ ($1000\text{ g}$) of the solvent. Unlike molarity, molality is independent of temperature because it is defined in terms of mass, which does not change with thermal fluctuations [Per the definition of concentration units in chemical thermodynamics].

2. Mathematical Expression

The molality of a solution is mathematically expressed as the ratio of the moles of solute to the mass of the solvent in kilograms:

$m = \frac{\text{Moles of solute} (n_B)}{\text{Mass of solvent in kilograms} (W_A \text{ in kg})}$

3. Derivation and Variable Breakdown

To determine the molality for a general binary solution containing solute $B$ and solvent $A$, we utilize the following variables:

• $n_B$: Number of moles of solute $= \frac{W_B}{M_B}$
• $W_B$: Mass of the solute in grams ($g$)
• $M_B$: Molar mass of the solute in grams per mole ($g \cdot mol^{-1}$)
• $W_A$: Mass of the solvent in grams ($g$)

Substituting these into the primary definition [By algebraic substitution]:

$m = \frac{W_B \times 1000}{M_B \times W_A (\text{in grams})}$

4. Conceptual Representation

5. Dimensional Analysis

Final Solution: Molality ($m$) is the concentration of a solution expressed as the number of moles of solute per kilogram of solvent, defined by the formula $m = \frac{n_B}{W_A(\text{kg})}$, with the SI unit $mol \cdot kg^{-1}$.

Theoretical Foundation: Principles of Solubility

The solubility of a solute in a solvent is governed by the empirical rule: "Like dissolves like." This principle relies on the nature of intermolecular forces. Solutes that exhibit similar polarity and magnitude of intermolecular forces as the solvent will dissolve readily [Per the Principle of Intermolecular Interaction].

• Solvent: $n\text{-octane} (CH_{3}(CH_{2})_{6}CH_{3})$ is a non-polar hydrocarbon.
• Solute-Solvent Interaction: Solutes with non-polar characteristics will exhibit strong dispersion (London) forces with $n\text{-octane}$, facilitating solubility. Polar or ionic solutes will have negligible interaction with the non-polar solvent, leading to poor solubility.

Step 1: Characterization of Solute Polarity

Step 2: Analyzing Interactions with $n\text{-octane}$

Solubility increases as the solute becomes more non-polar and capable of van der Waals interactions with the non-polar $n\text{-octane}$ chain.

1. Cyclohexane: Since both are non-polar hydrocarbons, they exhibit strong induced-dipole/induced-dipole interactions [London Dispersion Forces]. This results in maximum solubility.
2. $CH_{3}CN$: Contains a polar $-CN$ group, which disrupts the homogeneity of the non-polar solvent. Solubility is low.
3. $CH_{3}OH$: Features strong intermolecular hydrogen bonding. Energy required to disrupt these hydrogen bonds is not compensated by the non-polar solvent interaction [Enthalpic penalty], leading to very low solubility.
4. $KCl$: As an ionic solid, the lattice energy ($\Delta H_{lattice}$) is significantly higher than the energy released by solvation with non-polar molecules ($\Delta H_{solvation}$). It is effectively insoluble in $n\text{-octane}$.

Visual Representation of Polarity Gradient

Step 3: Final Ordering

Based on the progressive decrease in solute-solvent compatibility, the order of increasing solubility is:

$KCl < CH_{3}OH < CH_{3}CN < \text{Cyclohexane}$

Final Solution: The order of increasing solubility in n-octane is $KCl < CH_{3}OH < CH_{3}CN < \text{Cyclohexane}$. Solubility increases as the solute transitions from ionic to non-polar, maximizing interaction via London dispersion forces with the non-polar solvent.

Step 1: Calculating the Mass of Solute in Each Component

To determine the mass percentage of the final solution, we must first calculate the absolute mass of the solute contributed by each of the individual solutions. The formula for mass of solute is: $Mass_{solute} = \left( \frac{\text{Mass percentage}}{100} \right) \times \text{Total mass of solution}$ [Per the definition of mass percentage].

• For Solution A: $300\text{ g}$ of $25\%$ solution.
  $Mass_{solute, A} = 0.25 \times 300\text{ g} = 75\text{ g}$
• For Solution B: $400\text{ g}$ of $40\%$ solution.
  $Mass_{solute, B} = 0.40 \times 400\text{ g} = 160\text{ g}$

Step 2: Calculating Total Mass of Solute and Total Mass of Solution

Upon mixing the two solutions, the total mass of the solute and the total mass of the final solution are the sums of the individual components [Per the Principle of Conservation of Mass].

• Total Mass of Solute ($m_{total, solute}$):
  $75\text{ g} + 160\text{ g} = 235\text{ g}$
• Total Mass of Solution ($m_{total, solution}$):
  $300\text{ g} + 400\text{ g} = 700\text{ g}$

Step 3: Calculating Final Mass Percentage

The mass percentage of the resulting solution is calculated as follows:

$Mass\ \%\ = \left( \frac{\text{Total mass of solute}}{\text{Total mass of solution}} \right) \times 100$

$Mass\ \%\ = \left( \frac{235\text{ g}}{700\text{ g}} \right) \times 100$

$Mass\ \%\ = 0.335714... \times 100 \approx 33.57\%$

Visual Representation of Mixing Process

Final Solution: The mass percentage of the resulting solution is 33.57%.