Solutions – JEE Main ChemistryPractice Questions & PYQs

Osmotic pressure is a colligative property that depends on the total quantity of solute particles distributed throughout a specific volume. Because the product $\Pi V$ is proportional to the total moles of particles multiplied by $RT$, the overall pressure of a mixture can be determined by the sum of the individual pressure-volume contributions divided by the final combined volume.

For the sodium chloride solution, the product of its pressure and volume is $0.10 \text{ atm} \times 1 \text{ L} = 0.1 \text{ L atm}$. For the glucose solution, the contribution is $0.20 \text{ atm} \times 2 \text{ L} = 0.4 \text{ L atm}$. Combining these values and dividing by the total volume of $3 \text{ L}$ provides the final osmotic pressure according to the following equation:
$\Pi_{\text{final}} = \frac{0.1 + 0.4}{3} = \frac{0.5}{3} \text{ atm} \approx 0.1666 \text{ atm}$
Since this result is equivalent to $166.6 \times 10^{-3} \text{ atm}$, rounding to the nearest integer provides a value of $x = 167$.

Since the measured total vapour pressure of the mixture, 600 mm Hg, is greater than the vapour pressures of the pure components-512 mm Hg for $\mathrm{CS}_2$ and 344 mm Hg for acetone-the solution exhibits a positive deviation from Raoult's law. In such systems, the intermolecular attractive forces between the solute and solvent molecules are weaker than those between the individual components themselves, necessitating the absorption of heat to overcome these initial interactions. Because these repulsive-like interactions allow the molecules to expand into more space, the mixing process results in an increase in total volume, meaning a combination of 100 mL of $\mathrm{CS}_2$ and 100 mL of acetone will yield a total volume greater than 200 mL.

The temperature of the system is determined by the freezing point of the aqueous ethylene glycol solution, which undergoes freezing point depression relative to pure water. Using the relation $\Delta T_f = K_f \times m$, a 0.5 molal solution with a cryoscopic constant of $2.0\,\text{K kg mol}^{-1}$ results in a depression of $1.0\,\text{K}$. Subtracting this from the standard freezing point of water, $273\,\text{K}$, establishes the equilibrium temperature of the gas at $272\,\text{K}$.

Applying the ideal gas law $PV = nRT$ allows for the determination of the initial pressure of the gas within the cylinder. Substituting the values $n = 0.1\,\text{mol}$, $R = 0.08\,\text{dm}^3\,\text{atm K}^{-1}\,\text{mol}^{-1}$, $T = 272\,\text{K}$, and $V = 1.0\,\text{dm}^3$ yields an initial pressure $P_1$ of $2.176\,\text{atm}$. Upon the sudden withdrawal of the stoppers, the gas expands isothermally against an external atmospheric pressure of $1\,\text{atm}$. By equating the initial and final states using Boyle's law $P_1 V_1 = P_2 V_2$, where $P_2 = 1\,\text{atm}$, the final volume is found to be $2.176\,\text{litres}$.

When a solution is cooled below its freezing point, pure solvent separates as solid ice, causing the concentration of the remaining liquid solution to increase. This relationship is defined by the freezing point depression formula $\Delta T_f = K_f \times m$, where $m$ is the molality of the solution. Since the molar mass of ethylene glycol is 62 g/mol, the 62 g of solute equates to 1 mole. Given the freezing point depression of 10 K and the cryoscopic constant of 1.86 K kg mol⁻¹, the mass of the remaining liquid water $W_{kg}$ is calculated using the following equation:

$\Delta T_f = K_f \times \frac{1}{W_{kg}}$

$10 = 1.86 \times \frac{1}{W_{kg}}$

This calculation results in $W_{kg} = 0.186$ kg, which is equivalent to 186 g. Subtracting this from the initial mass of 250 g reveals that the mass of water separated as ice is $250 - 186 = 64$ g.

Henry's law establishes that the partial pressure of a gas above a solution, $P_{\text{gas}}$, is directly proportional to the mole fraction of the gas dissolved in the liquid, $X_{\text{gas}}$, expressed by the equation $P_{\text{gas}} = K_H \times X_{\text{gas}}$. Based on this relationship, the Henry's law constant $K_H$ is inversely proportional to the solubility of the gas, meaning that a larger $K_H$ value indicates that the gas is less soluble at a given pressure. Therefore, the claim that a higher value of $K_H$ leads to increased solubility directly contradicts this fundamental physical principle.

The boiling point elevation and freezing point depression are determined by the molality of the solution and the specific solvent constants, defined as $\Delta T_b = i \cdot m \cdot K_b$ and $\Delta T_f = i \cdot m \cdot K_f$. Since glucose is a non-electrolyte, its van 't Hoff factor is 1. Substituting the values for the boiling point elevation of 2 K in a 1 molal solution leads to $2 = 1 \cdot K_b$, which simplifies to $K_b = 2$. For the freezing point depression of 2 K in a 2 molal solution, we solve $2 = 2 \cdot K_f$, resulting in $K_f = 1$. Comparing these results indicates the relationship $K_b = 2 K_f$.

The osmotic pressure ($\pi$) of a solution is defined by the equation $\pi = iCRT$, where $i$ represents the van't Hoff factor, $C$ is the molar concentration, $R$ is the universal gas constant, and $T$ is the temperature. Complete dissociation of $XY$ into two ions ($X^+$ and $Y^-$) gives a van't Hoff factor $i = 2$, while complete dissociation of $\mathrm{BaCl}_2$ into three ions ($\mathrm{Ba}^{2+}$ and $2\mathrm{Cl}^-$) gives $i = 3$. Given that $\pi_{XY} = 4 \pi_{\mathrm{BaCl}_2}$ and assuming $R$ and $T$ are identical for both solutions, the relationship between the concentrations is $i_{XY} C_{XY} = 4 \times i_{\mathrm{BaCl}_2} C_{\mathrm{BaCl}_2}$. Substituting the known values leads to $2 \times C_{XY} = 4 \times (3 \times 0.01)$, which simplifies to $2 \times C_{XY} = 0.12$. Solving for the concentration results in $C_{XY} = 0.06 \text{ mol L}^{-1}$, which is equivalent to $6 \times 10^{-2} \text{ mol L}^{-1}$.

The van't Hoff factor, denoted by $i$, represents the ratio of observed particles to the number of formula units dissolved and relates to the degree of dissociation $\alpha$ through the equation $i = 1 + \alpha(n - 1)$. In this expression, $n$ stands for the total number of ions produced per formula unit of the solute. When $K_2HgI_4$ dissociates in water, it breaks apart into two potassium ions and one tetraiodomercurate(II) ion, which establishes the value $n = 3$. Given that the salt is 40% ionized, we use $\alpha = 0.4$ and calculate $i = 1 + 0.4(3 - 1)$, which simplifies to 1.8.

Raoult's Law establishes that the partial vapour pressure of a component in an ideal solution is the product of its liquid-phase mole fraction and its pure vapour pressure. Because the mole fraction of a component in the vapour phase is defined as the ratio of its partial pressure to the total pressure of the system, the vapour composition for the two components can be written as $y_M = \frac{x_M P_M^0}{P_t}$ and $y_N = \frac{x_N P_N^0}{P_t}$.

Comparing the mole fractions of the two components in the vapour phase by taking their ratio gives $\frac{y_M}{y_N} = \frac{x_M P_M^0}{x_N P_N^0}$. By substituting the specified pure vapour pressures of 450 mmHg for liquid $M$ and 700 mmHg for liquid $N$, the equation becomes $\frac{y_M}{y_N} = \frac{x_M}{x_N} \times \left( \frac{450}{700} \right)$. Since the factor $\frac{450}{700}$ is less than 1, the ratio in the vapour phase must be smaller than the ratio in the liquid phase to satisfy the equality. This leads to the relationship $\frac{x_M}{x_N} > \frac{y_M}{y_N}$.

Freezing point depression is a colligative property that depends on the total concentration of solute particles, so when benzoic acid dimerizes in benzene, we must account for the change in the number of particles using the Van't Hoff factor. For dimerization where $n = 2$ and the degree of association is 0.8, the factor is calculated as $i = 1 + \alpha(\frac{1}{n} - 1)$, resulting in $i = 1 + 0.8(\frac{1}{2} - 1) = 0.6$.

Using the freezing point depression formula $\Delta T_f = i K_f m$, we substitute the known values, where the molality $m$ is expressed as $\frac{w \times 1000}{122 \times 30}$. This gives the equation $2 = 0.6 \times 5 \times \frac{w \times 1000}{122 \times 30}$. Simplifying the expression results in $2 = 3 \times \frac{w \times 100}{122 \times 3}$, which reduces to $2 = \frac{w \times 100}{122}$. Rearranging to solve for $w$ gives $w = \frac{2 \times 122}{100}$, which equals 2.44 g.

Osmotic pressure is a colligative property that depends on the total concentration of solute particles, following the relationship $\pi = CRT$. When multiple non-electrolytes are present, the total osmotic pressure is the sum of their individual contributions, which is calculated based on the combined molarity of all dissolved particles in the solvent.

Determining the total concentration starts by calculating the molar amounts for each substance: urea provides $0.6/60 = 0.01$ moles, and glucose contributes $1.8/180 = 0.01$ moles. Because these solutes are dissolved in 100 mL of water, which is equivalent to 0.1 liters, the total molarity of the solution is $0.02/0.1 = 0.2$ moles per liter. Applying the osmotic pressure equation at a temperature of 300 Kelvin, the final calculation is $\pi = 0.2 \times 0.08206 \times 300$, resulting in 4.92 atm.

Henry's Law establishes that the partial pressure of a gas is directly proportional to its mole fraction in a solvent, a relationship mathematically defined as $P = K_H \cdot X$. When this data is represented graphically with partial pressure on the vertical axis and mole fraction on the horizontal axis, the Henry's law constant $K_H$ acts as the slope of the resulting straight line passing through the origin. Given the provided constants of 0.5 kbar for gas w, 2 kbar for gas x, 35 kbar for gas y, and 40 kbar for gas z, the steepness of the lines must follow the order w < x < y < z. The valid representation is the plot that shows these four distinct lines radiating from the origin, where the line for gas z exhibits the steepest gradient and the line for gas w appears the closest to the horizontal axis.

Determining the vapor composition of an ideal solution relies on the relationship where the mole fraction in the vapor phase is the ratio of a component's partial pressure to the total pressure of the mixture. Partial pressures are calculated using Raoult's Law, defined as $P_i = P_i^{\circ} x_i$, where $P_i^{\circ}$ is the pure vapor pressure and $x_i$ is the liquid mole fraction. For liquid A, the partial pressure is $7 \times 10^3 \times 0.4 = 2.8 \times 10^3$ Pa, and for liquid B, since its mole fraction is 0.6, the partial pressure is $12 \times 10^3 \times 0.6 = 7.2 \times 10^3$ Pa. Summing these partial pressures yields a total pressure of $10 \times 10^3$ Pa. The vapor phase mole fraction for each component is obtained by dividing its partial pressure by this total, resulting in $y_A = \frac{2.8 \times 10^3}{10 \times 10^3} = 0.28$ and $y_B = \frac{7.2 \times 10^3}{10 \times 10^3} = 0.72$.

Since the freezing point depression of an aqueous solution depends strictly on its molality, two solutions with identical freezing points must have equal molality. A 4% solution contains 4 grams of solute X per 96 grams of water, while a 12% solution of Y contains 12 grams per 88 grams of water. Equating the molality calculations for both gives:

$\frac{4/A}{96/1000} = \frac{12/M_Y}{88/1000}$

Solving for the molecular weight of Y leads to the following calculation:

$M_Y = \frac{12 \times 96 \times A}{4 \times 88} \approx 3.27A$

The value closest to this result is $3A$.

The elevation in boiling point for a solution is determined by the formula
$\Delta T_b = K_b \times m$
where $K_b$ represents the ebulliscopic constant of the solvent and $m$ denotes the molality of the solution. Since the mass of the solute and the solvent are identical for both scenarios, the molality remains constant, creating a direct proportionality between $\Delta T_b$ and $K_b$. Given that the ratio of the ebulliscopic constants for solvents A and B is $1:5$, the ratio of their boiling point elevations $\frac{\Delta T_b(A)}{\Delta T_b(B)}$ is also $1:5$.

According to Raoult's Law, the lowering of vapour pressure ($\Delta P$) for a solution containing a non-volatile solute is calculated using the equation $\Delta P = P^0 \times \chi_{solute}$, where $P^0$ is the vapour pressure of the pure solvent and $\chi_{solute}$ is the mole fraction of the solute. The number of moles of the solute and solvent are $n_{urea} = \frac{0.60}{60} = 0.01 \text{ mol}$ and $n_{water} = \frac{360}{18} = 20 \text{ mol}$. Substituting these values into the expression for lowering of vapour pressure yields:

$\Delta P = 35 \times \left( \frac{0.01}{0.01 + 20} \right)$
$\Delta P = 35 \times \frac{0.01}{20.01} \approx 0.01749 \text{ mmHg}$

This calculation results in a lowering of vapour pressure of approximately 0.017 mmHg.

Freezing point depression is a colligative property directly proportional to the concentration of the solute in the solution. Let $V_p$ be the volume of pure milk and $V_w$ be the volume of water added to form a total volume $V_p + V_w$. Since the freezing point depression is proportional to the concentration of the pure milk component, the relationship is expressed as:

$\frac{\Delta T_{diluted}}{\Delta T_{pure}} = \frac{V_p}{V_p + V_w}$

Substituting the observed values $0.2$ and $0.5$ for the respective depressions results in $0.2 / 0.5 = V_p / (V_p + V_w)$, which simplifies to $2 / 5 = V_p / (V_p + V_w)$. Cross-multiplying yields $5 V_p = 2 V_p + 2 V_w$, which simplifies to $3 V_p = 2 V_w$. This ratio indicates that 3 cups of water are present for every 2 cups of pure milk.

Because the final volume of the mixture is equal to the sum of the initial volumes, the solution behaves ideally and follows Raoult's Law. With the liquid phase mole fraction of B given as $X_B = 0.5$, the mole fraction of A is determined by $X_A = 1 - 0.5 = 0.5$. The total vapor pressure $P_T$ is calculated using the equation $P_T = P_A^0 X_A + P_B^0 X_B$. Substituting the provided vapor pressures of 400 mmHg and 600 mmHg, the total pressure is $P_T = (400 \times 0.5) + (600 \times 0.5) = 500$ mmHg.

The mole fractions in the vapor phase are determined using Dalton's Law, expressed by the formula $Y_i = \frac{P_i}{P_T}$, where the partial pressure $P_i$ is calculated as $P_i = P_i^0 X_i$. The partial pressure of component A is $400 \times 0.5 = 200$ mmHg. Applying this to the vapor phase equation results in $Y_A = \frac{200}{500} = 0.4$. Since the sum of the mole fractions in the vapor phase must equal 1, the mole fraction for component B is calculated as $Y_B = 1 - 0.4 = 0.6$.

Freezing point depression is a colligative property that depends on the total concentration of particles, requiring us to account for the van't Hoff factor when dealing with electrolytes. Because $\mathrm{K_2SO_4}$ dissociates completely into three ions, the van't Hoff factor is $i = 3$. Using the formula $\Delta T_f = i \times K_f \times m$ with the values $K_f = 4.0$ and $m = 0.03$, the calculation becomes $3 \times 4.0 \times 0.03 = 0.36$. This results in a final depression of 0.36 K.

Freezing point is a colligative property that depends on the total concentration of solute particles, where the freezing point $T_f$ is inversely proportional to the number of particles represented by the van 't Hoff factor $i$, often expressed as $T_f \propto \frac{1}{i}$. Because a smaller number of particles leads to less freezing point depression, the species that undergoes the least dissociation will maintain the highest freezing point. Comparing the given complexes, $[\mathrm{Co}(\mathrm{H_2O})_3\mathrm{Cl}_3]\cdot 3\mathrm{H_2O}$ is a neutral coordination compound that does not ionize in aqueous solution, resulting in the smallest van 't Hoff factor and consequently the highest freezing point.