Solutions – JEE Main ChemistryPractice Questions & PYQs

Raoult's Law for an ideal solution states that the total vapour pressure $P$ is given by $P = P^\circ_A x_A + P^\circ_B x_B$, where $P^\circ_A$ and $P^\circ_B$ are the vapour pressures of pure components A and B, and $x_A$ and $x_B$ are their respective mole fractions. For the first mixture containing 3 moles of A and 1 mole of B, the mole fractions are $x_A = 3/4$ and $x_B = 1/4$, leading to the following equation:
$500 = P^\circ_A \left( \frac{3}{4} \right) + P^\circ_B \left( \frac{1}{4} \right) \implies 3 P^\circ_A + P^\circ_B = 2000$
After adding 1 mole of A, the new composition is 4 moles of A and 1 mole of B, making the mole fractions $x_A = 4/5$ and $x_B = 1/5$. Given that the total pressure increases to 520 mm Hg, the equation becomes:
$520 = P^\circ_A \left( \frac{4}{5} \right) + P^\circ_B \left( \frac{1}{5} \right) \implies 4 P^\circ_A + P^\circ_B = 2600$
Subtracting the first equation from the second results in $P^\circ_A = 2600 - 2000 = 600$ mm Hg. Substituting this value back into the first equation yields $3(600) + P^\circ_B = 2000$, which simplifies to $P^\circ_B = 2000 - 1800 = 200$ mm Hg.

The mole fraction of component $A$ in the vapour phase $y_A$ is related to its mole fraction in the liquid phase $x_A$ by the expression
$y_A = \frac{x_A P_A^0}{x_A P_A^0 + (1 - x_A) P_B^0}$
Substituting the given values $y_A = 0.8$, $P_A^0 = 55$ kN m$^{-2}$, and $P_B^0 = 15$ kN m$^{-2}$ into the equation results in
$0.8 = \frac{55x_A}{55x_A + 15(1 - x_A)}$
Simplifying the denominator gives
$0.8 = \frac{55x_A}{40x_A + 15}$
Multiplying both sides by the denominator yields $32x_A + 12 = 55x_A$, which rearranges to $23x_A = 12$. Calculating the final value gives $x_A = \frac{12}{23} \approx 0.5217$.

Colligative properties like boiling point elevation and freezing point depression depend directly on the magnitudes of the ebullioscopic constant $K_b$ and the cryoscopic constant $K_f$. For water, the value of $K_b = 0.512 \ \mathrm{^{\circ}\mathrm{C} \ kg \ mol^{-1}}$ is lower than $K_f = 1.86 \ \mathrm{^{\circ}\mathrm{C} \ kg \ mol^{-1}}$, meaning that for any equimolal solution, the elevation in boiling point ($\Delta T_b$) must be smaller than the depression in freezing point ($\Delta T_f$). Consequently, osmotic pressure is the preferred technique for determining the molar masses of large molecules like proteins and polymers because it yields detectable values where other colligative properties produce changes that are too small to measure. Finally, the structural representation presented for the dimerised benzoic acid is inaccurate because it fails to correctly depict the hydrogen-bonded association between the carboxylic acid groups that characterizes the true dimer structure.

The van't Hoff factor quantifies how many particles a solute splits into when dissolved, which accounts for deviations from ideal behavior in colligative properties. Because the compound dissociates into three ions for each formula unit, the total number of particles per unit is $n = 3$. With a degree of dissociation of 30%, which is equivalent to $\alpha = 0.3$, we apply the relationship $i = 1 + (n-1)\alpha$ to determine the factor. Substituting these values into the expression yields $i = 1 + (3-1)(0.3)$, which simplifies to $i = 1 + 0.6 = 1.6$. Representing this final value as required, we obtain $16 \times 10^{-1}$.

Freezing point depression, represented by $\Delta T_f$, occurs because the presence of a non-volatile solute lowers the vapor pressure of a liquid solvent. On a diagram plotting vapor pressure against temperature, the equilibrium curve for the liquid solution sits below the curve for the pure solvent, while the vapor pressure curve for the solid solvent remains unaffected. The freezing point is identified as the temperature where the solid phase vapor pressure curve intersects the liquid curve; therefore, the lowered solution curve inevitably leads to an intersection at a lower temperature compared to the pure solvent. This specific shift, showing the intersection with the solution line occurring to the left of the solvent line along the temperature axis, illustrates the concept correctly.

Adding a solute like NaCl to water induces a colligative property known as the depression of freezing point, which causes the freezing point of the mixture to drop significantly below the standard $0^{\circ}\,\mathrm{C}$ of pure water. This combination creates a freezing mixture, generating an environment cold enough to effectively keep ice cream frozen or prevent it from melting. Since the addition of salt physically lowers the freezing point of the ice, both statements are factually correct.

Freezing point depression is a colligative property governed by the relationship $\Delta T_f = K_f \cdot m$, where $\Delta T_f$ represents the decrease in temperature and $m$ denotes the molality of the solution. Determining the cryoscopic constant requires finding the molality by dividing the moles of solute by the mass of the solvent in kilograms. With 1 g of solute having a molar mass of 256 g mol$^{-1}$, the amount of substance is $1/256$ moles, and the 50 g of solvent equates to 0.050 kg. This results in a molality of $(1/256) / 0.050$, which simplifies to $1000 / 12800$ mol kg$^{-1}$. Applying the equation $K_f = \Delta T_f / m$ using the given depression of 0.40 K yields $K_f = 0.40 / (1000 / 12800)$, resulting in 5.12 K kg mol$^{-1}$.

The lowering of vapour pressure for a solution containing a non-volatile solute is directly proportional to the mole fraction of that solute, represented by $\Delta P \propto x_{\text{solute}}$. Given that a vapour pressure decrease of 10 mm of Hg corresponds to a solute mole fraction of 0.2, doubling the decrease to 20 mm of Hg requires the solute's mole fraction to also double, reaching 0.4. Since the sum of the mole fractions of the solute and solvent in any binary solution is always 1, the mole fraction of the solvent is obtained by subtracting the solute's mole fraction from 1. Subtracting 0.4 from 1 gives the final mole fraction of the solvent as 0.6.

Boiling point elevation is a colligative property that depends on the number of solute particles in a solution, described by the relationship $\Delta T_b = K_b \times m$, where the molality $m$ represents moles of solute per kilogram of solvent. By rearranging this expression to isolate molar mass, we can determine the mass of one mole of a compound using the formula $\text{Molar mass} = \frac{K_b \times w}{\Delta T_b \times W}$, where $w$ is the mass of the solute in grams and $W$ is the mass of the solvent in kilograms.

For the first compound AB, substituting the elevation of $2.7 \text{ K}$ and the solvent mass of $0.015 \text{ kg}$ into the formula yields a molar mass of $\frac{0.5 \times 1}{2.7 \times 0.015}$, which is $12.35 \text{ g/mol}$. Applying the same calculation for $\text{AB}_2$ with an elevation of $1.5 \text{ K}$ results in $\frac{0.5 \times 1}{1.5 \times 0.015}$, giving a molar mass of $22.22 \text{ g/mol}$.

Designating the atomic masses of A and B as $a$ and $b$ allows for the creation of the system of equations $a + b = 12.35$ and $a + 2b = 22.22$. Subtracting the first equation from the second isolates the value of $b$ as $9.87 \text{ g/mol}$, and substituting this back into the first equation leads to $a = 2.48 \text{ g/mol}$. Writing this result in the required format gives $24.8 \times 10^{-1} \text{ amu}$, which rounds to the nearest integer 25.

The elevation in boiling point is a colligative property directly proportional to the product of the van't Hoff factor and the solute concentration, governed by the relation $\Delta T_{b}=i K_{b} m$. First, identify the van't Hoff factor ($i$) for each solute, noting that $i = 2$ for $\text{NaCl}$ due to its dissociation into two ions, while $i = 1$ for urea as it does not dissociate. Multiplying these factors by the molar concentrations yields the relative particle concentration for each solution: urea at $10^{-4} \mathrm{M}$ results in $1 \times 10^{-4}$, $\text{NaCl}$ at $10^{-4} \mathrm{M}$ results in $2 \times 10^{-4}$, $\text{NaCl}$ at $10^{-3} \mathrm{M}$ results in $2 \times 10^{-3}$, and $\text{NaCl}$ at $10^{-2} \mathrm{M}$ results in $2 \times 10^{-2}$. Comparing these resulting values directly establishes the increasing order of boiling points as (ii) < (i) < (iii) < (iv).

The molal depression constant $K_f$ is fundamentally related to the enthalpy of fusion and the freezing temperature of the solvent by the equation $K_f = \frac{R T_f^2 M_1}{1000 \Delta H_{fus}}$. Utilizing the thermodynamic definition of the molar entropy of fusion, where $\Delta S_{fus} = \frac{\Delta H_{fus}}{T_f}$, substituting $\Delta H_{fus} = T_f \Delta S_{fus}$ into the equation validates the theoretical expression for $K_f$ given in Statement I.

Regarding the specific values for these solvents, the experimental $K_f$ for benzene is approximately $5.12 \, \text{K kg mol}^{-1}$, whereas the $K_f$ for water is approximately $1.86 \, \text{K kg mol}^{-1}$. Because the value for benzene is significantly higher than that for water, the assertion in Statement II that the $K_f$ for benzene is less than the $K_f$ for water is incorrect.

Reverse osmosis occurs when a pressure $p_1$ greater than the osmotic pressure $\pi$ is applied to the concentrated solution side (chamber 1), effectively forcing solvent molecules to move against the osmotic gradient toward the dilute solution (chamber 2). This phenomenon strictly requires the use of a semi-permeable membrane, such as cellophane or parchment paper, which selectively permits the passage of solvent molecules while blocking the solute. Porous membranes are incapable of supporting this process because they fail to block solute particles, and applying pressure to the dilute side $p_2$ is insufficient to induce the necessary reverse flow.

According to Raoult's Law, the total vapour pressure of a mixture of volatile liquids relates to the vapour pressure of its pure components and their mole fractions, as described by the formula $P_{\text{total}} = P_A^{\circ} x_A + P_B^{\circ} x_B$. Given 1 mole of A and 3 moles of B, the mole fraction for A is 0.25 and for B is 0.75. By applying these values to the law, the equation becomes $500 = 200(0.25) + P_B^{\circ}(0.75)$, which simplifies to $500 = 50 + 0.75 P_B^{\circ}$. Isolating the unknown variable results in $450 = 0.75 P_B^{\circ}$, leading to a pure vapour pressure for B of 600 mm Hg. Because the least volatile component is defined by the lowest pure vapour pressure, and 200 mm Hg is lower than 600 mm Hg, A is identified as the least volatile component.

The Henry's Law constant, $K_H$, determines the solubility of a gas in water at a specific pressure, such as 1 bar, and its relationship with temperature follows a distinct, non-linear trend. Starting at low temperatures, $K_H$ increases as the temperature rises, eventually reaching a maximum value. Beyond this peak, the value of $K_H$ begins to decrease, creating a characteristic curve that accurately represents this temperature-dependent behavior.

Minimum boiling azeotropes arise in liquid mixtures that exhibit a significant positive deviation from Raoult's law, characterized by weaker intermolecular attractions between unlike molecules than those between identical molecules. Because these weaker forces result in higher vapor pressure, the mixture boils at a temperature lower than either of its components. Conversely, mixtures showing negative deviation from Raoult's law form maximum boiling azeotropes due to stronger intermolecular forces, which is the specific behavior of the mixture of $\mathrm{C}_6\mathrm{H}_5\mathrm{OH}$ and $\mathrm{C}_6\mathrm{H}_5\mathrm{NH}_2$.

Boiling point elevation occurs because the presence of non-volatile solute particles hinders the solvent from vaporizing, thereby requiring a higher temperature to reach the boiling point. Since both ethylene glycol and glucose are non-electrolyte solutes, the total quantity of solute is the sum of 2 moles and 2 moles, which equals 4 moles. Given the solvent mass is 500 g, or 0.5 kg, the molality is found by dividing the moles of solute by the mass of the solvent in kilograms, resulting in $\frac{4}{0.5} = 8 \text{ mol kg}^{-1}$. Using the formula $\Delta T_b = K_b \times m$, the temperature increase is calculated as $\Delta T_b = 0.52 \times 8 = 4.16 \text{ K}$. Adding this value to the boiling point of pure water, 373.15 K, leads to a final boiling point of 377.3 K.

Osmosis drives solvent molecules from a region of lower solute concentration to one of higher concentration across a semi-permeable membrane. A cell containing $0.9\% (\text{w/w})$ glucose is inherently hypotonic compared to the surrounding medium defined in the problem. The composition of this external solution is set by equal mole fractions of glucose and water, represented as $X_{\text{glucose}} = 0.5$ and $X_{\text{water}} = 0.5$.

Determining the mass of this mixture requires using the molar masses of glucose, $180\,\text{g/mol}$, and water, $18\,\text{g/mol}$. For equal mole fractions, the mass contributed by glucose is $0.5 \times 180 = 90\,\text{g}$ and the mass of water is $0.5 \times 18 = 9\,\text{g}$, which results in a total solution weight of $99\,\text{g}$. The weight percentage concentration is thus $\frac{90}{99} \times 100 \approx 90.9\% (\text{w/w})$. Because the external solution is drastically more concentrated than the cell fluid, water will flow out of the cell, causing it to shrink, which signifies that none of the specific numerical options provided matches this physical reality.

The elevation of the boiling point is a colligative property defined by the relationship $\Delta T_{\mathrm{b}} = i K_{\mathrm{b}} m$. For the first compound $AX_2$, the molality is $m = \frac{1.24}{124 \times 1} = 0.01$. Applying this to the observed boiling point elevation $\Delta T_{\mathrm{b}} = 0.0156$ results in the equation $0.0156 = i \times 0.52 \times 0.01$, which simplifies to $i = 3$. Since the dissociation of $AX_2$ into three ions follows the expression $i = 1 + 2\alpha$, the calculated value confirms that $\alpha = 1$, meaning $AX_2$ is fully ionized.

For the second compound $AY_2$, the molality is calculated as $m = \frac{25.4}{250 \times 2} = 0.0508$. Substituting these values into the boiling point elevation formula, we obtain $0.0260 = i \times 0.52 \times 0.0508$. Solving for $i$ yields $i \approx 1$. Because the Van't Hoff factor matches the value expected for a non-electrolyte, $AY_2$ is completely unionized.

When analyzing the behavior of binary liquid solutions, identifying the nature of intermolecular forces and their deviations from ideal behavior reveals how the mixture interacts. Chloroform mixed with acetone forms a solution that exhibits a negative deviation from Raoult's Law, creating stronger intermolecular attractions than the pure components and resulting in a maximum boiling azeotrope. Conversely, a mixture of ethanol and water displays a positive deviation from Raoult's Law as the components disrupt each other's hydrogen bonding, which leads to a minimum boiling azeotrope.

Solutions composed of benzene and toluene act as ideal solutions because the intermolecular forces between the different molecules are virtually the same as those between identical molecules, ensuring that $\Delta V_{mix} = 0$ throughout the mixing process. As a distinct chemical phenomenon in non-polar solvents, acetic acid molecules engage in hydrogen bonding to form pairs, or dimerize, when dissolved in benzene.

In an ideal mixture, the more volatile component-the one with the higher pure vapor pressure-contributes more significantly to the vapor phase, causing the vapor composition to differ from the liquid composition. Since the pure vapor pressures are $350\ \mathrm{mmHg}$ for $A$ and $750\ \mathrm{mmHg}$ for $B$, the vapor phase is naturally enriched in component $B$. Combining Raoult's Law with Dalton's Law establishes the relationship $\frac{y_A}{y_B} = \frac{P_A^0 X_A}{P_B^0 X_B}$, which simplifies to $\frac{y_A}{y_B} = \frac{350}{750} \left( \frac{X_A}{X_B} \right)$. Because the ratio $\frac{350}{750}$ is less than one, the vapor phase mole fraction ratio $\frac{y_A}{y_B}$ must be smaller than the liquid phase mole fraction ratio $\frac{X_A}{X_B}$.