Solutions – JEE Main ChemistryPractice Questions & PYQs

Henry's law states that the partial pressure of a gas above a liquid is directly proportional to its mole fraction in the solution, expressed as $P_{\mathrm{gas}} = K_{\mathrm{H}} \cdot \chi_{\mathrm{gas}}$. In scenarios involving dilute solutions, the mole fraction is effectively equal to the ratio of the moles of gas to the moles of solvent. With 0.9 L of water weighing 900 g, we determine the solvent moles using $n_{\mathrm{H}_2\mathrm{O}} = \frac{900}{18} = 50$ mol.

Substituting the given partial pressure and Henry's constant into the relation leads to:
$0.835 = 1.67 \times 10^{3} \cdot \frac{n_{\mathrm{CO}_2}}{50}$
Rearranging to isolate the unknown quantity yields $n_{\mathrm{CO}_2} = \frac{0.835 \times 50}{1.67 \times 10^{3}}$, which results in 0.025 mol. Converting this value into millimoles by multiplying by 1000 gives $x = 25$.

The behavior of dilute solutions is characterized by the van't Hoff equation, which correlates osmotic pressure with temperature and molar concentration. This principle is defined by the formula $\pi = CRT$, where $\pi$ is the osmotic pressure, $C$ is the molarity, $R$ is the universal gas constant, and $T$ is the absolute temperature. Because molarity is equivalent to the mass of the solute divided by the product of its molar mass and the volume of the solution, we rearrange the definition to isolate the molar mass $M$ using the equation $M = \frac{wRT}{\pi V}$.

Substituting the known values, where the mass $w$ is 1.46 g, the volume $V$ is 0.1 L, the temperature $T$ is 300 K, and the osmotic pressure $\pi$ is 2.42 x 10^-3 bar, leads to the following calculation:
$M = \frac{1.46 \times 0.083 \times 300}{2.42 \times 10^{-3} \times 0.1}$
Solving this equation provides a molar mass of 15.02 multiplied by 10^4 g mol^-1. Rounding this figure to the nearest integer results in 15.

The principle of freezing point depression states that the decrease in the freezing point of a solution, denoted by $\Delta T_f$, is directly proportional to the Van't Hoff factor, the molal depression constant, and the molality of the solution, expressed by the relationship $\Delta T_f = i \times K_f \times m$. For a weak acid that dissociates in water, the Van't Hoff factor $i$ accounts for the increased number of particles and is related to the degree of dissociation $\alpha$ by the formula $i = 1 + \alpha$.

Given the freezing point depression value of $3.885$, the molal depression constant of $1.85$, and the molality of $2$, we substitute these values into the fundamental equation to determine the Van't Hoff factor:
$3.885 = i \times 1.85 \times 2$
Simplifying the right side of the equation results in $3.885 = i \times 3.7$, which leads to $i = \frac{3.885}{3.7} = 1.05$. By substituting this result into the dissociation expression $1 + \alpha = 1.05$, we find that $\alpha = 0.05$. Converting this decimal value to the required scientific notation gives $50 \times 10^{-3}$, which corresponds to the degree of dissociation of the acid.

Raoult's Law and Dalton's Law allow us to determine the vapor phase composition by first finding the partial pressures of the components in the liquid phase. Since the sum of mole fractions in a mixture equals 1, a mixture with a mole fraction of A of 0.6 must have a mole fraction of B of 0.4, leading to partial pressures of $P_A = 0.6 \times 90 = 54$ mm Hg and $P_B = 0.4 \times 15 = 6$ mm Hg. Summing these gives a total vapor pressure of $P_T = 54 + 6 = 60$ mm Hg, which allows us to find the mole fraction of B in the vapor phase using the ratio $Y_B = 6 / 60 = 0.1$. As this result can be expressed as $1 \times 10^{-1}$, the value of $x$ is 1.

Freezing point depression occurs when a non-volatile solute is added to a solvent, lowering the temperature at which the liquid solidifies, a process governed by the equation $\Delta T_f = K_f \times m$. Because ethylene glycol does not dissociate in water, its van't Hoff factor is 1, which allows us to determine the freezing point reduction based solely on the concentration of the solute.

To find the concentration, first calculate the molar mass of ethylene glycol ($\mathrm{C_2H_6O_2}$), which is 62 g/mol. The 83 g of solute corresponds to 1.339 moles, and dividing this by the solvent mass of 0.625 kg gives the molality, calculated using the equation $m = 1.339 / 0.625$, which equals 2.142 mol/kg. Applying the depression constant through the equation $\Delta T_f = 1.86 \times 2.142$ yields a temperature decrease of approximately 3.98 K. Subtracting this value from the original freezing point of water, 273 K, results in a final freezing point of 269 K.

Freezing point depression is a colligative property, meaning the degree to which the freezing point of a solvent drops depends on the total number of dissolved particles rather than the specific nature of the solute. This relationship is defined by the equation $\Delta T_f = i K_f m$, where $\Delta T_f$ is the freezing point depression, $i$ is the van't Hoff factor representing the number of particles produced per formula unit, $K_f$ is the freezing point depression constant, and $m$ is the molality. Because the freezing point of a solution is determined by subtracting this depression from the freezing point of the pure solvent, solutions with a higher concentration of solute particles exhibit a larger $\Delta T_f$ and consequently a lower freezing point.

The solution of 0.10 M $\mathrm{C}_2\mathrm{H}_5\mathrm{OH}$ consists of non-dissociating molecules, resulting in a van't Hoff factor of 1 and an effective particle concentration of 0.10 M. Conversely, all four of the listed salt solutions-$\mathrm{Ba}_3(\mathrm{PO}_4)_2$, $\mathrm{Na}_2\mathrm{SO}_4$, $\mathrm{KCl}$, and $\mathrm{Li}_3\mathrm{PO}_4$-are electrolytes that dissociate into multiple ions when dissolved in water. Because these salts break apart into more than one particle per formula unit, their van't Hoff factors are all strictly greater than 1, which elevates their effective solute concentration well above the 0.10 M concentration found in the ethanol solution. Since every one of these four electrolyte solutions provides a higher concentration of solute particles than the ethanol solution, they all undergo a greater magnitude of freezing point depression, resulting in all four of these solutions possessing a freezing point lower than that of the 0.10 M $\mathrm{C}_2\mathrm{H}_5\mathrm{OH}$ solution.

Colligative properties like boiling point elevation depend entirely on the total concentration of dissolved particles in a solution. Since the two solutions share the same boiling point, they must necessarily contain the same total effective molality.

A 1 molal solution of $\mathrm{K}_4\mathrm{Fe}(\mathrm{CN})_6$ dissociates into five ions: four potassium ions and one ferrocyanide complex ion. Given a degree of dissociation of 0.4, the total effective molality is calculated by adding the concentration of undissociated salt, which is 0.6 m, to the concentration of the resulting ions, 1.6 m and 0.4 m, totaling 2.6 m. Consequently, the non-electrolytic solute A must produce this same effective molality of 2.6 m to match the boiling point of the first solution.

An 18.1 weight percent solution contains 18.1 g of solute in 81.9 g of solvent. Using the definition of molality with the mass of the solvent expressed in kilograms, the molar mass $M$ is found from the following equation:
$2.6 = \frac{18.1 / M}{0.0819}$
Solving for $M$ results in a molar mass of 85.

Colligative properties depend on the actual concentration of particles in a solution, necessitating the use of the van't Hoff factor to account for the dissociation of electrolytes. For a binary compound like AB that splits into two ions, the van't Hoff factor is determined by $i = 1 + \alpha(n - 1)$, where $\alpha$ represents the degree of dissociation and $n$ is the number of ions produced per molecule. Given a dissociation of 75 percent, the factor is calculated to be 1.75.

The boiling point elevation is governed by the equation $\Delta T_b = i \cdot K_b \cdot m$. Rearranging this to solve for molality gives $m = \frac{\Delta T_b}{i \cdot K_b}$. Substituting the provided values of 2.5 for the rise in boiling point, 0.52 for the ebullioscopic constant, and 1.75 for the van't Hoff factor results in a calculation of 2.5 divided by the product of 1.75 and 0.52. This division yields approximately 2.74, which rounds to the nearest integer of 3 molal.

Freezing point depression describes how the addition of a solute lowers the temperature at which a solvent solidifies, and in this system, cooling causes pure water to turn into ice, increasing the concentration of the remaining liquid. A 1 kg total mass allows us to express the initial amount of water as $w_w$ grams and sucrose as $1000 - w_w$ grams. By defining molality as $0.75 = \frac{(1000 - w_w)/342}{w_w/1000}$, we isolate $w_w$ to determine that the system starts with approximately 795.86 g of water and 0.5969 moles of sucrose.

Cooling the solution to $-4^{\circ}\ \text{C}$ creates a freezing point depression $\Delta T_{\mathrm{f}}$ of $4^{\circ}\ \text{C}$, triggering water to crystallize until the remaining solvent reaches a new equilibrium. Utilizing the formula $\Delta T_{\mathrm{f}} = K_{\mathrm{f}} \times m$, we can substitute the known values into $4 = 1.86 \times \frac{0.5969}{w_{new}}$, where $w_{new}$ is the mass of the remaining liquid water in kilograms. Solving for $w_{new}$ yields approximately 0.2775 kg, or 277.5 g. The amount of ice separated is the difference between the starting mass of water and this final liquid mass, resulting in $795.86 - 277.5 = 518.36\ \text{g}$, which rounds to 518 g.

Boiling point elevation is a colligative property proportional to the molality of the solution and the van't Hoff factor of the solute, defined by the equation $\Delta T_b = i K_b m$. Because the mass of the organic acid and the mass of the solvent are kept constant in both experimental setups, the molality remains the same, which allows for a direct comparison between the two systems. In the acetone solution, the acid acts as a monomer with a van't Hoff factor of 1, and with 1.7 K kg mol$^{-1}$ as the ebullioscopic constant, the boiling point elevation is 0.17 °C. When dissolved in benzene, the acid dimerizes, which effectively reduces the van't Hoff factor to 0.5 while the ebullioscopic constant is 2.6 K kg mol$^{-1}$.

Dividing the expression for benzene by the expression for acetone eliminates the common molality terms, resulting in the relationship $\frac{\Delta T_{b(\text{benzene})}}{\Delta T_{b(\text{acetone})}} = \frac{0.5 \times 2.6}{1 \times 1.7}$. Substituting the known elevation for acetone into this ratio gives $\frac{\Delta T_{b(\text{benzene})}}{0.17} = \frac{1.3}{1.7}$. Solving for the elevation in benzene leads to $\Delta T_{b(\text{benzene})} = 0.17 \times \frac{1.3}{1.7}$, which simplifies to 0.13 °C. Representing this result as $13 \times 10^{-2}$, the value of x is 13.

Henry's Law states that the partial pressure of a gas above a solvent is directly proportional to its solubility within the solution, represented by the equation $p = K_H \chi$. Given a partial pressure of 20 kPa and a Henry's law constant of $8.0 \times 10^4$ kPa, the solubility is determined by the ratio $\chi = \frac{p}{K_H}$. Substituting these values, the expression $\frac{20}{8.0 \times 10^4}$ results in $2.5 \times 10^{-4}$. Adjusting this figure to the specified format, $2.5 \times 10^{-4}$ becomes $25 \times 10^{-5}$, yielding 25.

Osmosis is the governing principle here, describing the net movement of water molecules through a semi-permeable membrane from a region of higher water concentration to one of lower concentration. Since the salt solution is significantly more concentrated than the internal fluid of the raw mango, water molecules are drawn out of the fruit to balance these osmotic pressure differences. This continuous loss of internal fluid is what causes the mango to lose volume and shrink.

A pure solvent possesses a higher vapor pressure than a solution containing a non-volatile solute, a relationship defined by $P_{solution} < P_{solvent}^\circ$. Because the pure solvent exerts a higher vapor pressure within the sealed container, its molecules continuously evaporate into the headspace to establish equilibrium. Consequently, these solvent vapors condense into the solution beaker to match its lower equilibrium pressure, causing a net transfer of liquid that increases the volume of the solution while decreasing the volume of the pure solvent.

The principle of freezing point depression dictates that the reduction in the freezing temperature of a solution is directly proportional to its molality, taking into account the number of ions produced by the solute. For an ionic compound like sodium chloride, we use the van't Hoff factor $i = 2$ to account for its dissociation into sodium and chloride ions. We apply the formula $\Delta T_f = i \times K_f \times m$, where $\Delta T_f$ is the depression, $K_f$ is the cryoscopic constant, and $m$ is the molality.

With a freezing point depression of 0.2 K and a constant $K_f$ of 2 K kg mol$^{-1}$ for 600 g of water, we express molality as the moles of NaCl, $w / 58.5$, divided by the mass of the solvent in kilograms, 0.6 kg. Substituting these values into the depression equation gives $0.2 = 2 \times 2 \times \frac{w / 58.5}{0.6}$. Simplifying this mathematical expression leads to $0.2 = \frac{4w}{35.1}$, which yields a mass $w$ of 1.755 g of NaCl to achieve the target freezing point.

Two solutions are isotonic when they exert the same osmotic pressure, a condition that occurs when their molar concentrations are identical because the temperature remains constant. Osmotic pressure is calculated using the equation $\pi = \frac{w}{MV}RT$, where $w$ is the mass in grams, $M$ is the molar mass in grams per mole, $V$ is the volume in liters, $R$ is the gas constant, and $T$ is the temperature in Kelvin.

By equating the osmotic pressure of protein A and protein B, we can write $\frac{0.73}{M_A \times 0.25}RT = \frac{1.65}{M_B \times 1}RT$. Since the temperature of 298 K and the gas constant are the same for both systems, these terms cancel out, simplifying the relationship to $\frac{0.73}{0.25 M_A} = \frac{1.65}{M_B}$. Rearranging this equation to solve for the ratio of the molecular masses of A and B gives $\frac{M_A}{M_B} = \frac{0.73}{0.25 \times 1.65}$, which results in approximately 1.77. This corresponds to a value of 177 when expressed as $177 \times 10^{-2}$.

The relative lowering of vapour pressure for a dilute solution is directly determined by the mole fraction of the non-volatile solute, which follows the relationship $\frac{\Delta P}{P^{0}} = x_{\text{solute}}$. Because the mass of the water solvent is identical in all three cases, we find its molar quantity is 10 moles by dividing the 180 g mass by the molar mass of 18 g mol⁻¹.

Each solute's contribution to the mole fraction depends on its number of moles, which is calculated as the mass divided by the molar mass. Solute A provides 0.1 moles, solute B contributes 0.05 moles, and solute C yields 0.001 moles. Since the relative lowering of vapour pressure is proportional to the number of moles of solute when the amount of solvent remains constant, we compare these values. The resulting molar amounts follow the sequence $0.1 > 0.05 > 0.001$, which establishes that the relative lowering of vapour pressure is greatest for solute A and least for solute C, resulting in the order A > B > C.

The vapour pressure of a liquid is inversely related to the strength of its intermolecular forces, which can be summarized by the relationship $P_{vap} \propto \frac{1}{\text{Intermolecular Forces}}$. Since the graph indicates that liquid X maintains a higher vapour pressure than liquid Y at any given temperature, it follows that X must possess weaker intermolecular interactions than Y. This confirms that inference (B) is accurate, as the physical properties of the liquids demonstrate that Z possesses the strongest intermolecular forces, followed by Y, and finally X.

The elevation of boiling point is given by the formula $\Delta T_b = i \cdot K_b \cdot m$, where $i$ is the van't Hoff factor, $K_b$ is the ebullioscopic constant, and $m$ is the molality. Given that the elevation of boiling point for the complex solution is twice that of the $\mathrm{CaCl}_2$ solution, the relationship can be expressed as $i_1 \cdot m_1 \cdot K_b = 2 \cdot (i_2 \cdot m_2 \cdot K_b)$, where $m_1 = 0.10 \, \mathrm{m}$ and $m_2 = 0.05 \, \mathrm{m}$. Since $\mathrm{CaCl}_2$ dissociates into three ions ($\mathrm{Ca}^{2+}$ and $2\mathrm{Cl}^-$), its van't Hoff factor $i_2$ is $3$, which leads to $i_1 \cdot 0.10 = 2 \cdot (3 \cdot 0.05)$, solving to $i_1 = 3$.

For the complex represented by the formula $\mathrm{CrCl}_3 \cdot x\mathrm{NH}_3$, it can be written in the coordination form $[\mathrm{Cr}(\mathrm{NH}_3)_x \mathrm{Cl}_y] \mathrm{Cl}_z$, where the number of ions $i_1 = 1 + z = 3$, implying $z = 2$ ionizable chloride atoms. Given that the coordination number of chromium is $6$ and all $\mathrm{NH}_3$ molecules are present inside the coordination sphere, the sum of the ligands is $x + y = 6$. Since the total number of chlorine atoms provided by the complex is $y + z = 3$, substituting $z = 2$ gives $y = 1$. Finally, substituting $y = 1$ into the coordination sphere equation $x + y = 6$ yields $x = 5$.

Henry's Law states that the partial pressure of a gas is directly proportional to its mole fraction in a solution, expressed by the relationship $P = K_H X$. For a 55.5 molal solution, there are 55.5 moles of solute present for every 1 kilogram of solvent. Since 1 kilogram of water contains approximately $\frac{1000}{18} \approx 55.5$ moles, the mole fraction $X$ for the solute is calculated as $\frac{55.5}{55.5 + 55.5}$, which equals 0.5.

For gas $\delta$, the given Henry's constant is $K_H = 0.5\,\text{kbar}$, which is equivalent to 500 bar. Substituting these values into the Henry's law equation yields $P_{\delta} = 500\,\text{bar} \times 0.5 = 250\,\text{bar}$. This demonstrates that a 55.5 molal solution of gas $\delta$ exerts a pressure of 250 bar.

Raoult's Law defines the total vapour pressure of an ideal solution as the sum of the individual partial vapour pressures, which are calculated by multiplying the pure vapour pressure of each component by its respective mole fraction in the mixture. Initially, with 1 mole of n-hexane and 3 moles of n-heptane, the solution consists of 4 moles total, resulting in an expression for total pressure of $550 = P_{hexane}^o \times \frac{1}{4} + P_{heptane}^o \times \frac{3}{4}$, which simplifies to $P_{hexane}^o + 3P_{heptane}^o = 2200$.

When an additional mole of n-heptane is introduced, the total number of moles becomes 5, with 1 mole of n-hexane and 4 moles of n-heptane, leading to a total vapour pressure of 560 mm of Hg. This second condition is represented by $560 = P_{hexane}^o \times \frac{1}{5} + P_{heptane}^o \times \frac{4}{5}$, which simplifies to $P_{hexane}^o + 4P_{heptane}^o = 2800$. By subtracting the initial equation from this new equation, the unknown term for n-hexane cancels out, directly yielding $P_{heptane}^o = 2800 - 2200 = 600$ mm of Hg.