Solutions – JEE Main ChemistryPractice Questions & PYQs

Gibbs free energy is a fundamental thermodynamic state function defined by the equation $G = H - TS$, which quantifies the maximum non-expansion work obtainable from a system at constant temperature and pressure. When a system transitions between two distinct solution states, the alteration in composition or internal configuration necessarily shifts its chemical potential. This fundamental change in the thermodynamic profile ensures that the total Gibbs free energy of the system must vary during the process.

The extent to which a weak acid dissociates directly determines the observed freezing point depression of its solution, as the total number of particles increases with dissociation. We begin by finding the Van 't Hoff factor, $i$, which represents the ratio of observed solute particles to the theoretical amount, using the relationship $i = \frac{\Delta T_f}{K_f m}$. Substituting the provided values, we calculate $i = \frac{0.20}{1.8 \times 0.1}$, resulting in $i = 1.11$. Since the monobasic acid HA dissociates into two ions, the relationship between the degree of dissociation $\alpha$ and the Van 't Hoff factor is simplified to $i = 1 + \alpha$, which gives $\alpha = 0.11$. Finally, by applying the equilibrium constant expression $K_a = \frac{c \alpha^2}{1 - \alpha}$ with a concentration $c = 0.1$, we compute $K_a = \frac{0.1 \times (0.11)^2}{1 - 0.11}$, leading to the value $1.38 \times 10^{-3}$.

Molarity is defined by the formula $M = \frac{n}{V}$, where $n$ represents the constant amount of dissolved NaCl and $V$ represents the volume of the solution. Since water exhibits anomalous expansion, its volume decreases as it is heated from 1°C to 4°C, causing the denominator of the molarity expression to shrink and the concentration to rise to a maximum value at 4°C. As the temperature continues to increase from 4°C to 25°C, the water expands and the volume increases, which leads to a corresponding decrease in the molarity.

Molality ($m$) is defined as the number of moles of solute per kilogram of solvent. For $1\ \text{L}$ of sea water with a density of $2\ \text{g mL}^{-1}$, the total mass of the solution is $2000\ \text{g}$. Given a concentration of $6\ \text{M}$ for NaCl, the mass of the solute is $6\ \text{mol} \times 58.5\ \text{g mol}^{-1} = 351\ \text{g}$, making the mass of the solvent (water) $2000\ \text{g} - 351\ \text{g} = 1649\ \text{g}$.

The oxygen concentration of $5.8\ \text{ppm}$ implies there are $5.8\ \text{g}$ of $\mathrm{O}_2$ in $10^6\ \text{g}$ of solution. The mass of the solvent in this amount of solution is $\frac{1649\ \text{g}}{2000\ \text{g}} \times 10^6\ \text{g} = 824500\ \text{g} = 824.5\ \text{kg}$. Calculating the moles of $\mathrm{O}_2$ as $\frac{5.8\ \text{g}}{32\ \text{g mol}^{-1}} = 0.18125\ \text{mol}$, the molality is $m = \frac{0.18125\ \text{mol}}{824.5\ \text{kg}} \approx 2.198 \times 10^{-4}\ \text{m}$. Rounding $2.198$ to the nearest integer, $x = 2$.

Raoult's Law dictates that the total pressure of a binary solution containing two volatile components is the sum of their partial pressures, expressed as $P_T = P_1^0 x_1 + P_2^0 x_2$. Since the sum of mole fractions in the liquid phase is unity, $x_2 = 1 - x_1$, allowing the total pressure to be written as $P_T = P_1^0 x_1 + P_2^0(1 - x_1)$, which simplifies to $P_T = x_1(P_1^0 - P_2^0) + P_2^0$. Given that the vapor phase mole fraction is $y_1 = \frac{P_1^0 x_1}{P_T}$, taking the reciprocal yields $\frac{1}{y_1} = \frac{P_T}{P_1^0 x_1}$. Substituting the expanded form of $P_T$ into this equation results in $\frac{1}{y_1} = \frac{x_1(P_1^0 - P_2^0) + P_2^0}{P_1^0 x_1}$, which further reduces to $\frac{1}{y_1} = \frac{P_1^0 - P_2^0}{P_1^0} + \frac{P_2^0}{P_1^0} \left( \frac{1}{x_1} \right)$. Rearranging this expression to isolate $\frac{1}{x_1}$ gives $\frac{P_2^0}{P_1^0} \left( \frac{1}{x_1} \right) = \frac{1}{y_1} - \frac{P_1^0 - P_2^0}{P_1^0}$. Multiplying the entire equation by $\frac{P_1^0}{P_2^0}$ leads to the linear form $\frac{1}{x_1} = \left( \frac{P_1^0}{P_2^0} \right) \frac{1}{y_1} + \frac{P_2^0 - P_1^0}{P_2^0}$, where the slope is $\frac{P_1^0}{P_2^0}$ and the intercept is $\frac{P_2^0 - P_1^0}{P_2^0}$.

Positive deviations from Raoult's Law occur in mixtures where the intermolecular forces between the two components are weaker than the attractions within the pure substances themselves, making it easier for molecules to escape into the vapor phase. In the mixture of $(\mathrm{CH}_3)_2\mathrm{CO}$ and $\mathrm{CS}_2$, the interaction between the polar acetone and the non-polar carbon disulfide is significantly weaker than the cohesive forces found within the pure components, which prevents the molecules from being held as tightly in the liquid state. This resulting reduction in intermolecular attraction causes the total vapor pressure of the solution to be higher than what would be predicted for an ideal system.

Osmotic pressure depends on the total concentration of solute particles present on either side of a semipermeable membrane, and the net pressure arises from the difference in these concentrations. When electrolytes are present, the total concentration must account for the complete dissociation of the solute into individual ions.

Inside the artificial cell, glucose remains as individual molecules, maintaining a concentration of 0.2 M. Outside the cell, NaCl dissociates completely into sodium and chloride ions, so the 0.05 M concentration of NaCl results in an effective particle concentration of 0.1 M. The net concentration difference driving the osmosis is 0.1 M, and applying the osmotic pressure formula gives:

$\pi = (0.2 - 0.1) \times 0.083 \times 300 = 2.49$

Converting this to the requested unit format results in 24.9 x 10^-1 bar, which rounds to the nearest integer of 25.

The freezing point of a solution decreases compared to the pure solvent because the addition of a solute lowers its chemical potential, a colligative property known as freezing point depression. When a weak acid like acetic acid is dissolved, it undergoes partial ionization, which increases the total number of solute particles in the solution; this necessitates the use of a Van't Hoff factor to accurately calculate the shift in the freezing point.

The mass of the dissolved acetic acid is $5\ \mathrm{mL} \times 1.2\ \mathrm{g/mL}$, resulting in $6\ \mathrm{g}$ of solute. Given the molar mass of $60\ \mathrm{g/mol}$, this mass corresponds to $0.1\ \mathrm{mol}$. Because this amount is dissolved in $1\ \mathrm{L}$ of water, which weighs $1\ \mathrm{kg}$ at a density of $1\ \mathrm{g/cm^3}$, the molality of the solution is $0.1\ \mathrm{m}$. The degree of dissociation $\alpha$ follows from the equilibrium relation $\alpha = \sqrt{\frac{K_a}{C}}$ using $K_a = 6.25 \times 10^{-5}$ and concentration $C = 0.1\ \mathrm{M}$, which yields $\alpha = 0.025$. Consequently, the Van't Hoff factor $i = 1 + \alpha$ equals $1.025$. Applying the freezing point depression formula $\Delta T_f = i \times K_f \times m$ with the given $K_f = 1.86\ \mathrm{K\ kg\ mol^{-1}}$ results in $\Delta T_f = 1.025 \times 1.86 \times 0.1 = 0.19065\ {}^{\circ}\mathrm{C}$. Representing this value as $19 \times 10^{-2}\ {}^{\circ}\mathrm{C}$ indicates that $x = 19$.

When two liquids exhibit a negative deviation from Raoult's law, the intermolecular forces of attraction between the mixed components are significantly stronger than the forces within the pure components. This increased attraction restricts the escape of molecules into the vapor phase, resulting in a total vapor pressure that is lower than the value predicted for an ideal solution, expressed as $P_{T} < P_{A^0} X_{A} + P_{B^0} X_{B}$. Because the boiling point is defined as the temperature at which vapor pressure reaches atmospheric pressure, this reduction in vapor pressure necessitates a higher temperature to reach that threshold, which directly leads to an increase in the boiling point.

When two substances are mixed, the component present in the greater molar amount acts as the solvent, while the other serves as the solute. Comparing the molar amounts reveals that $2.7 \, \text{kg}$ of water corresponds to $150$ moles, while $2.7 \, \text{kg}$ of acetic acid corresponds to $45$ moles, identifying water as the solvent. The freezing point depression is governed by the relation $\Delta T_f = K_f \times m$, where $K_f$ is the cryoscopic constant for the solvent and $m$ is the molality of the solution. Because the solute neither dissociates nor dimerizes, the van't Hoff factor is taken as one.

Molality represents the moles of solute per kilogram of solvent. Dividing $45$ moles of acetic acid by $2.7 \, \text{kg}$ of water yields a molality of $\frac{45}{2.7} \, \text{mol} \, \text{kg}^{-1}$. Applying the constant $K_f = 1.86 \, \text{K} \, \text{kg} \, \text{mol}^{-1}$ for water, the depression in freezing point becomes $\Delta T_f = 1.86 \times \frac{45}{2.7} = 31 \, \text{K}$. Since pure water freezes at $0^\circ \text{C}$, the freezing point of the solution is $-31^\circ \text{C}$, establishing $x = 31$.

Freezing point depression is a colligative property governed by the relationship $\Delta T_f = K_f m$, which describes how the freezing point of a solvent decreases in proportion to the molality of the solute. Because adding methanol to water lowers the freezing point, the change in temperature is calculated as $\Delta T_f = 273.15 - 270.65 = 2.5 \text{K}$. Since the solution consists of 100 mL of water, which corresponds to 0.1 kg of solvent, the molality is the number of moles $n$ divided by 0.1. Substituting the known values into the equation gives $2.5 = 1.86 \times \frac{n}{0.1}$, which simplifies to $n = 0.1344 \text{ moles}$.

Determining the mass of the methanol involves multiplying these moles by the molar mass, resulting in $w = 0.1344 \times 32 = 4.3 \text{g}$. To find the volume of the added methanol, divide this mass by the given density, yielding $V = \frac{4.3}{0.792} = 5.43 \text{mL}$. Expressing this volume as a product of $10^{-2} \text{mL}$ gives $543 \times 10^{-2} \text{mL}$, identifying $x$ as 543.

The depression in freezing point of a solvent depends directly on the molality of the dissolved solute, which is expressed by the relationship:
$\Delta T_f = K_f \times m$
Because ethylene glycol does not dissociate in water, the van't Hoff factor is taken as 1. The target temperature depression is 24 K, and the freezing point constant for water is 1.86 K kg mol^-1.

Molality represents the moles of solute per kilogram of solvent, calculated here by dividing the mass of ethylene glycol in grams by its molar mass of 62 g mol^-1, and then dividing by the solvent mass of 18.6 kg. Substituting these values into the formula gives:
$24 = 1 \times 1.86 \times \frac{W}{62 \times 18.6}$
Simplifying the expression by noting that 18.6 divided by 1.86 equals 10 results in:
$24 = \frac{W}{620}$
Solving for the mass W results in 14880 g, which converts to 14.88 kg and rounds to a final value of 15 kg.

The van 't Hoff factor ($i$) measures the effective number of particles in an electrolyte solution, and for a strong electrolyte like $\mathrm{NaCl}$, this value trends toward the theoretical limit of 2 as the concentration decreases. Higher concentrations lead to stronger interionic attractions that restrict the independent movement of ions and lower the effective particle count, while increasing dilution weakens these attractions, resulting in a higher observed $i$ value. Specifically, the values of 1.87 for 0.1 M, 1.94 for 0.01 M, and the continued approach toward 2 for 0.001 M demonstrate that the factor increases as the solution becomes more dilute, establishing the order $i_A < i_B < i_C$.

The freezing point of a solvent like benzene is a colligative property, meaning it changes based on the presence of dissolved solute particles. When naphthalene is added to benzene, it acts as a non-volatile solute that disrupts the solvent's ability to crystallize, which is quantitatively described by the relationship $\Delta T_f = K_f \cdot m$. Because the presence of the solute lowers the freezing point compared to the pure solvent, the mixture undergoes a depression in freezing point, resulting in a lower temperature requirement for solidification.

The elevation in boiling point, described by the formula $\Delta T_b = K_b \cdot m$, allows us to determine the molality of the solution given the elevation of 2 °C and the constant $K_b = 0.52 \, \text{K} \cdot \text{kg} \cdot \text{mol}^{-1}$, which yields a molality of $m = \frac{2}{0.52} \, \text{mol} \cdot \text{kg}^{-1}$. Since the solution is highly dilute, we apply Raoult's Law for the relative lowering of vapor pressure, expressed as $\frac{\Delta P}{P^{\circ}} = \frac{m \cdot M_{\text{solvent}}}{1000}$, using the molar mass of water $M_{\text{solvent}} = 18 \, \text{g} \cdot \text{mol}^{-1}$ and an initial vapor pressure $P^{\circ} = 760 \, \text{mm} \, \text{Hg}$. Substituting these values, the reduction in vapor pressure is $\Delta P = 760 \times \frac{(2/0.52) \times 18}{1000} = 52.615 \, \text{mm} \, \text{Hg}$. Subtracting this change from the initial pressure, the vapor pressure of the resulting solution is $\Delta P_{sol} = 760 - 52.615 = 707.385 \, \text{mm} \, \text{Hg}$, which rounds to 707 mm Hg.

Osmosis occurs when two solutions of different concentrations are separated by a semi-permeable membrane, driving solvent molecules to move spontaneously from the region of lower solute concentration to the region of higher solute concentration. Since the $\mathrm{CuSO}_4$ solution has a higher concentration of 0.05 M compared to the 0.01 M $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ solution, water molecules will migrate through the membrane into the $\mathrm{CuSO}_4$ side. This influx of solvent increases the total volume of the $\mathrm{CuSO}_4$ solution without changing the number of moles of the solute, which mathematically lowers its overall molarity.

Osmotic pressure behaves analogously to gas pressure in an ideal solution, following the relationship $\pi = CRT$. Because the concentration of the solution remains constant, the osmotic pressure is directly proportional to the absolute temperature, which allows us to relate the two states using the ratio $\frac{\pi_1}{T_1} = \frac{\pi_2}{T_2}$.

By rearranging this relationship to solve for the unknown final pressure, we obtain $\pi_2 = \frac{\pi_1 T_2}{T_1}$. Substituting the initial pressure of $7 \times 10^5 \text{ Pa}$ at $273 \text{ K}$ and the final temperature of $283 \text{ K}$, we get $\pi_2 = \frac{7 \times 10^5 \times 283}{273}$. Converting the initial pressure to $70 \times 10^4 \text{ Pa}$ allows the final result to be expressed in the required units, leading to $\frac{70 \times 10^4 \times 283}{273}$, which yields a value of $72.56 \times 10^4 \text{ Nm}^{-2}$.

Raoult's Law describes how the partial pressure of each component in an ideal solution is calculated by multiplying its pure vapor pressure by its mole fraction in the liquid state. Since the mixture is equimolar, the liquid mole fraction for both benzene and methyl benzene is $x = 0.5$. Calculating the partial pressures provides 40 Torr for benzene and 12 Torr for methyl benzene, leading to a total pressure of $40 + 12 = 52$ Torr. The mole fraction of methyl benzene in the vapor phase is determined by the ratio of its partial pressure to the total pressure of the system, represented by the equation $y = \frac{12}{52}$. Performing this calculation results in 0.23, which is equivalent to $23 \times 10^{-2}$.

The depression in freezing point is determined by the expression $\Delta T_f = i \cdot K_f \cdot m$, which indicates that the solution with the largest calculated value for $\Delta T_f$ will exhibit the lowest freezing point. For $180 \mathrm{g}$ of glucose dissolved in water, the molality is $1 \mathrm{mol/kg}$, and with $i=1$ and the cryoscopic constant $K_f = 1.86 \mathrm{K \cdot kg \cdot mol^{-1}}$, the depression is $1 \times 1.86 \times 1 = 1.86^{\circ} \mathrm{C}$. When $180 \mathrm{g}$ of benzoic acid is dissolved in benzene, the molality is approximately $1.48 \mathrm{mol/kg}$, and because benzoic acid dimerizes in benzene, $i=0.5$; using the cryoscopic constant for benzene $K_f = 5.12 \mathrm{K \cdot kg \cdot mol^{-1}}$, the depression becomes $0.5 \times 5.12 \times 1.48 = 3.8^{\circ} \mathrm{C}$. Dissolving $180 \mathrm{g}$ of acetic acid in water produces a molality of $3 \mathrm{mol/kg}$ with $i \approx 1$, resulting in a depression of $1 \times 1.86 \times 3 = 5.58^{\circ} \mathrm{C}$. Finally, $180 \mathrm{g}$ of acetic acid in benzene results in a molality of $3 \mathrm{mol/kg}$, and due to dimerization in this solvent, $i=0.5$, which yields a depression of $0.5 \times 5.12 \times 3 = 7.68^{\circ} \mathrm{C}$. Since $7.68^{\circ} \mathrm{C}$ is the largest value among all scenarios, the acetic acid solution in benzene produces the highest depression in freezing point.

The elevation in boiling point for an electrolyte solution depends on the total concentration of dissolved particles, which is determined by the van 't Hoff factor, $i$. This factor adjusts the ideal boiling point elevation equation to $\Delta T_b = i K_b m$, where $K_b$ represents the molal boiling point elevation constant and $m$ is the molality of the solution. For the electrolyte $\mathrm{AB}_2$, which dissociates into one cation and two anions, the van 't Hoff factor is expressed by the relationship $i = 1 + 2\alpha$, where $\alpha$ is the degree of ionization.

The elevation in boiling point is the difference between the boiling point of the solution and the solvent, calculated as $100.52^\circ\mathrm{C} - 100^\circ\mathrm{C} = 0.52\,\mathrm{K}$. To find the molality, we calculate the moles of solute as $10\,\mathrm{g} / 200\,\mathrm{g}\,\mathrm{mol}^{-1} = 0.05\,\mathrm{mol}$ and divide by the mass of the solvent, $100\,\mathrm{g}$, which is $0.1\,\mathrm{kg}$, resulting in $m = 0.5\,\mathrm{mol}\,\mathrm{kg}^{-1}$. Substituting these values into the boiling point elevation formula, we obtain $0.52 = i \times 0.52 \times 0.5$, which simplifies to $i \times 0.5 = 1$, or $i = 2$. By substituting this into the dissociation expression $1 + 2\alpha = 2$, we determine that $2\alpha = 1$, yielding $\alpha = 0.5$, which is equivalent to 5 multiplied by $10^{-1}$.