Solutions – JEE Main ChemistryPractice Questions & PYQs

The elevation in boiling point is a colligative property defined by the relation $\Delta T_{b} = i \cdot K_{b} \cdot m$. Because sodium chloride dissociates into two ions while urea is a non-electrolyte, their respective van 't Hoff factors differ, meaning the boiling point elevation cannot be identical for both solutions at the same concentration. Azeotropic mixtures behave distinctively as they boil without changing composition, implying the concentration of components in the liquid phase is identical to that in the vapor phase. Osmosis is the spontaneous movement of solvent from a region of lower solute concentration, known as a hypotonic solution, to a region of higher solute concentration, termed a hypertonic solution, which contradicts the provided statement. Regarding the density of the $32\% \text{ H}_2\text{SO}_4$ solution, applying the formula $M = \frac{10 \cdot \% \cdot d}{MW}$ with the given molarity of $4.09 \text{ M}$ confirms the density is approximately $1.26 \text{ g mL}^{-1}$. Finally, when potassium iodide is added to silver nitrate, the silver nitrate is in excess, causing the resulting silver iodide colloidal particles to adsorb silver ions from the dispersion medium, which produces a positively charged sol rather than a negatively charged one.

Two solutions are defined as isotonic when they exert the same osmotic pressure at a given temperature, requiring their product of van't Hoff factor and molarity to be equal. For a non-electrolyte like glucose, the van't Hoff factor is $1$, so equating the two solutions gives $0.01 = i(0.004)$, resulting in $i = 2.5$. Because $\mathrm{K}_2\mathrm{SO}_4$ dissociates into $n = 3$ ions, we use the expression $i = 1 + (n - 1)\alpha$ to find that $2.5 = 1 + (3 - 1)\alpha$, which simplifies to $2\alpha = 1.5$ and yields a degree of dissociation of $0.75$, or $75\%$.

Two solutions are defined as isotonic when they exert the same osmotic pressure at a given temperature, a condition determined by the concentration of dissolved particles. The governing relationship for osmotic pressure is $\pi = icRT$, where $i$ represents the van't Hoff factor, $C$ is the molarity, $R$ is the ideal gas constant, and $T$ is temperature. Because temperature and the gas constant are identical for all solutions, comparing osmotic pressures reduces to checking if the product of the van't Hoff factor and molarity, $iC$, is equal for each pair.

In the first pair, $1\ \mathrm{M}$ $\mathrm{NaCl}$ has $i=2$ and $\mathrm{Urea}$ is a non-electrolyte with $i=1$, yielding $iC$ products of $2 \times 1 = 2$ and $1 \times 2 = 2$, respectively, which are equal. The second pair consists of $1\ \mathrm{M}$ $\mathrm{CaCl}_2$ with $i=3$ and $1.5\ \mathrm{M}$ $\mathrm{KCl}$ with $i=2$, producing $iC$ values of $3 \times 1 = 3$ and $2 \times 1.5 = 3$. For the third pair, $1.5\ \mathrm{M}$ $\mathrm{AlCl}_3$ has $i=4$ and $2\ \mathrm{M}$ $\mathrm{Na}_2\mathrm{SO}_4$ has $i=3$, resulting in equal $iC$ products of $6$. Finally, $2.5\ \mathrm{M}$ $\mathrm{KCl}$ with $i=2$ and $1\ \mathrm{M}$ $\mathrm{Al}_2(\mathrm{SO}_4)_3$ with $i=5$ both result in an $iC$ value of $5$. Since every pair satisfies the requirement of having equal $iC$ products, all four pairs are isotonic.

Raoult's Law dictates that the presence of a non-volatile solute reduces the vapour pressure of a solvent, with the relative lowering being directly proportional to the solute's mole fraction. Establishing the solution's composition reveals that a 30% (w/v) concentration means 30 g of glucose exists in 100 mL of solution. Since the density is 1.2 g cm⁻³, the solution weighs 120 g, leaving the mass of the water solvent at 90 g. Molality is determined by dividing the moles of glucose, which are $30 \text{ g} / 180 \text{ g mol}^{-1} \approx 0.1667 \text{ mol}$, by the solvent mass of 0.09 kg, resulting in $1.85 \text{ mol kg}^{-1}$. Applying the relation $\frac{P^{\circ} - P_s}{P_s} = \frac{m \cdot M_{\text{solvent}}}{1000}$ with a pure water vapour pressure $P^{\circ}$ of 24 mm Hg and solvent molar mass of 18 g mol⁻¹ provides $\frac{24 - P_s}{P_s} = \frac{1.85 \times 18}{1000}$, which simplifies to approximately 0.0333. Rearranging to solve for $P_s$ leads to $24 = 1.0333 P_s$, yielding a final vapour pressure of 23 mm Hg.

Freezing point depression is a colligative property that is directly proportional to the total concentration of solute particles in a solution. Because acetic acid is a weak electrolyte that undergoes partial dissociation, the effective number of particles in the solution is greater than the number of undissociated molecules, which is accounted for by the van't Hoff factor $i = 1 + \alpha$, where $\alpha$ represents the degree of dissociation.

To determine the amount of acetic acid, calculate the total mass by multiplying the volume by the density: $1.2\ \mathrm{mL} \times 1.02\ \mathrm{g\ mL}^{-1} = 1.224\ \mathrm{g}$. Dividing this mass by the molar mass of $60\ \mathrm{g\ mol}^{-1}$ yields $0.0204\ \mathrm{mol}$. Given that the solution volume is $2.0\ \mathrm{L}$, the molality of the solution is $0.0204\ \mathrm{mol} / 2.0\ \mathrm{kg} = 0.0102\ \mathrm{mol\ kg}^{-1}$. Applying these values to the freezing point depression equation $\Delta T_f = i \times K_f \times m$, we get $0.0198 = (1 + \alpha) \times 1.85 \times 0.0102$. Solving for $\alpha$ results in $0.04928$, which corresponds to a dissociation of approximately $4.928\%$. Rounded to the nearest integer, the percentage of dissociation is $5$.

Molality is defined as the number of moles of solute per kilogram of solvent, represented by $m = \frac{n_{\text{solute}}}{w_{\text{solvent(kg)}}}$, and since mass remains invariant regardless of temperature shifts, this concentration term is inherently temperature-independent. Unlike molarity, which depends on volume that contracts or expands when cooling a 5 M KCl solution from 10 °C to -21 °C, the physical mass of the solute and solvent stays exactly the same. Consequently, because mass does not fluctuate with thermal conditions, the molality remains constant, which confirms that the stated reason is the correct explanation for the assertion.

Raoult's Law establishes that the relative lowering of vapour pressure in a solution is equal to the mole fraction of the non-volatile solute. In a dilute solution, this is represented as $\frac{P^0 - P_S}{P^0} = \frac{n_{\text{solute}}}{n_{\text{solvent}}}$, where $P^0$ is the pure solvent vapour pressure and $P_S$ is the solution vapour pressure. Although the pure water pressure is defined as $23.76\ \text{mmHg}$, this specific constant cancels out in the ratio calculations.

Since the vapour pressure is reduced to one-half of the pure water value, the equation becomes $\frac{P^0 - 0.5P^0}{P^0} = 0.5$, which indicates the mole fraction of the solute is $0.5$. The number of moles of the solvent is found by dividing the mass of $100\ \text{g}$ of water by the molar mass of $18\ \text{g/mol}$, resulting in approximately $5.56\ \text{mol}$.

Establishing the relationship $\frac{n_{\text{solute}}}{n_{\text{solvent}}} = 0.5$ leads to the number of moles of solute being $5.56 \times 0.5$, which equals $2.78\ \text{mol}$. Rounding this value to the nearest integer results in 3.

For a glucose solution to be isotonic with blood, it must exert the same osmotic pressure, a relationship governed by the van't Hoff equation $\pi = CRT$, where $C$ is the molar concentration, $R$ is the gas constant, and $T$ is the temperature. By substituting the known values into the equation $7.47 = C \times 0.083 \times 300$, we can solve for $C$ to determine the molarity of the solution in moles per liter. Multiplying this molarity by the molar mass of glucose, 180 g/mol, converts the concentration into the required mass concentration in g/L, which results in 54 g/L.

Determining the composition of the vapor phase requires finding the ratio of a specific component's partial pressure to the total pressure of the system. According to Raoult's law, the partial pressure of each liquid is the product of its pure vapor pressure and its mole fraction in the liquid mixture, which allows us to define the total pressure as the sum of these individual contributions. Given that the mole fraction of liquid A is 0.3, the mole fraction of liquid B is 0.7. Calculating the partial pressures yields $P_A = 50 \times 0.3 = 15$ Torr and $P_B = 100 \times 0.7 = 70$ Torr, resulting in a total pressure of $P_{total} = 15 + 70 = 85$ Torr. Finally, the mole fraction of component B in the vapor phase is determined by the ratio $y_B = \frac{P_B}{P_{total}} = \frac{70}{85}$, which simplifies to $\frac{14}{17}$, meaning the value of $x$ is 14.

Osmotic pressure follows the Van 't Hoff equation $\pi = CRT$, which relates pressure to molar concentration, the universal gas constant, and absolute temperature. The concentration $C$ is determined by dividing the mass of the protein, $2.0\ \text{g}$, by its molar mass of $60,000\ \text{g}\ \text{mol}^{-1}$ and the solution volume of $0.2\ \text{L}$. Given the gas constant $R = 0.083\ \text{L}\ \text{bar}\ \text{mol}^{-1}\ \text{K}^{-1}$ and a temperature of $300\ \text{K}$, substituting these values into the osmotic pressure equation yields:

$\pi = \frac{2.0\ \text{g}}{60000\ \text{g}\ \text{mol}^{-1} \times 0.2\ \text{L}} \times 0.083\ \text{L}\ \text{bar}\ \text{mol}^{-1}\ \text{K}^{-1} \times 300\ \text{K}$

Evaluating this expression results in a pressure of $0.00415\ \text{bar}$. Because the final value is required in Pascals, applying the conversion factor $1\ \text{bar} = 10^5\ \text{Pa}$ provides the final result of $415\ \text{Pa}$.

The freezing point depression of an aqueous solution is governed by the total number of solute particles present, which for an electrolyte is determined by the van 't Hoff factor. Since potassium chloride dissociates into two ions in water, the observed freezing point depression is greater than what would be expected for a non-electrolyte of the same concentration.

To calculate the percentage dissociation, first determine the molality of the 0.5 percent potassium chloride solution, which contains 0.5 g of solute in 99.5 g of solvent. The molality is calculated by dividing the moles of solute (0.5 g / 74.6 g/mol) by the mass of the solvent in kilograms (0.0995 kg). Using the relationship $\Delta T_{f} = i K_{f} m$, we rearrange to solve for the van 't Hoff factor: $i = \frac{\Delta T_{f}}{K_{f} m}$ Substituting the given values into this expression yields $i = \frac{0.24}{1.80 \times \left( \frac{0.5}{74.6 \times 0.0995} \right)} = 1.98$ Finally, apply the formula for the degree of dissociation $\alpha$, which is $\alpha = \frac{i-1}{n-1}$ Given that $n=2$ for potassium chloride, this becomes $\alpha = \frac{1.98-1}{2-1} = 0.98$. This value corresponds to 98 percent dissociation.

Boiling point elevation is a colligative property defined by the expression $\Delta T_b = K_b m$, meaning that if two aqueous solutions of non-volatile solutes exhibit the same boiling point, they must have identical molalities. A 2% solution of solute A consists of 2 grams of solute in 98 grams of water, giving a molality calculated as $m_A = \frac{2}{M_A \cdot 0.098}$. An 8% solution of solute B contains 8 grams of solute in 92 grams of water, resulting in a molality of $m_B = \frac{8}{M_B \cdot 0.092}$. Equating these two values, $\frac{2}{0.098 M_A} = \frac{8}{0.092 M_B}$, isolates the relationship between the two molecular weights. Rearranging the variables leads to $\frac{M_B}{M_A} = \frac{8 \cdot 0.098}{2 \cdot 0.092}$, which evaluates to approximately 4.26, indicating that $M_B = 4 M_A$ provides the best approximation among the choices.

The freezing point depression of a solution is a colligative property determined by the total number of particles present, which changes when solute molecules associate. By evaluating the ratio of the observed freezing point depression to the theoretical depression calculated for a non-associating solute, we determine the van't Hoff factor $i$, which quantifies the extent of this association.

To find $i$, we first calculate the molality $m$ of the solution using the provided mass of the solute and the mass of the solvent, giving $m = \frac{0.7\,\mathrm{g}}{93\,\mathrm{g}\,\mathrm{mol}^{-1}} \times \frac{1000\,\mathrm{g}}{42.0\,\mathrm{g}}$. Substituting the depression constant $K_f = 1.86\,\mathrm{K}\,\mathrm{kg}\,\mathrm{mol}^{-1}$ and the freezing point depression $\Delta T_f = 0.2^{\circ}\,\mathrm{C}$ into the equation $\Delta T_f = i K_f m$, we set up $0.2 = i \times 1.86 \times (\frac{0.7}{93} \times \frac{1000}{42.0})$, which yields $i = 0.60$.

For the dimerization process $2A \rightleftharpoons A_2$, the van't Hoff factor relates to the degree of association $\alpha$ through the expression $i = 1 - \alpha + \frac{\alpha}{2}$. Setting this expression equal to the calculated value of $0.60$, we obtain $1 - \frac{\alpha}{2} = 0.60$, which simplifies to $\frac{\alpha}{2} = 0.40$, resulting in a degree of association of $\alpha = 0.80$, or $80\%$.

Henry's law provides the governing principle for this problem, establishing that the partial pressure of a gas is directly proportional to its mole fraction in solution, as expressed by the equation:
$P = K_H \cdot X$
In this context, $X$ represents the mole fraction of the dissolved oxygen. Because the solubility is minimal, this fraction can be simplified to the ratio of the moles of oxygen to the moles of water, $X \approx \frac{n_{\mathrm{O}_2}}{n_{\mathrm{H}_2\mathrm{O}}}$.

Considering one litre of water, which has a mass of 1000 g, the number of moles of solvent is $\frac{1000}{18} \approx 55.55$ mol. By substituting the given values of $P = 0.920$ bar and $K_H = 46.82 \times 10^3$ bar into the rearranged formula $n_{\mathrm{O}_2} = \frac{P \times n_{\mathrm{H}_2\mathrm{O}}}{K_H}$, the expression becomes:
$n_{\mathrm{O}_2} = \frac{0.920 \times 1000}{46.82 \times 10^3 \times 18}$
This calculation simplifies to $n_{\mathrm{O}_2} = \frac{0.920}{46.82 \times 18} \approx 0.00109$ mol. Converting this result to millimoles yields approximately 1.09 mmol, which corresponds to 1 when rounded to the nearest integer.

The elevation in boiling point is directly proportional to the ebullioscopic constant of the solvent, governed by the equation
$\Delta T_b = K_b \cdot m$
where $m$ is the molality. Because the mass of the solute and the mass of the solvent are identical for both systems, the molality remains constant, meaning the ratio of boiling point elevations corresponds directly to the ratio of the ebullioscopic constants. Since the ratio of the ebullioscopic constants for the two solvents is $1:8$, the ratio of the boiling point elevations $\frac{x}{y}$ is equal to $\frac{1}{8}$, which establishes that $y = 8$.

Henry's law establishes the relationship between partial pressure $p$ and mole fraction $x$ through the equation $p = K_H \cdot x$. Since the mole fraction is consistent across the experimental points shown, the partial pressure $p$ is directly proportional to the Henry's law constant $K_H$, meaning that higher values of $K_H$ result in higher partial pressures.

Comparing the provided constants, gas A possesses the largest $K_H$ value of 30.40 kbar, while gas C features the smallest value of 1.56 x 10^-5 kbar. Because line 'p' represents the maximum partial pressure and line 's' signifies the minimum, they correspond to gas A and gas C, respectively.

Colligative properties such as the elevation in boiling point and the depression in freezing point are directly proportional to the molality of the solute, defined by the relationships $\Delta T_{b} = K_{b} m$ and $\Delta T_{f} = K_{f} m$. A 1 molal solution of solute A with an elevation in boiling point of 3 K establishes that $K_{b}$ equals 3 since $3 = K_{b} \times 1$. Similarly, a 2 molal solution of A showing a depression in freezing point of 6 K indicates that $K_{f}$ also equals 3 because $6 = K_{f} \times 2$. Consequently, the ratio $\frac{K_{b}}{K_{f}}$ simplifies to 1. Since this ratio is represented as $\frac{1}{X}$, it follows that X is 1.

Colligative properties such as boiling point elevation and freezing point depression are directly proportional to the molality of a solute, defined by the formulas
$\Delta T_b = i K_b m$
and
$\Delta T_f = i K_f m$
Since glucose is a non-electrolyte, its van 't Hoff factor $i = 1$, meaning the changes in temperature depend solely on the molality multiplied by the respective constants $K_b$ and $K_f$.

With the provided conditions, the boiling point elevation equation is $4 = K_b \times 1.5$ and the freezing point depression equation is $4 = K_f \times 4.5$. Because both temperature changes are equal, we can relate the two equations by setting them equal to each other, resulting in $K_b \times 1.5 = K_f \times 4.5$. Solving for the ratio $\frac{K_b}{K_f}$ by dividing 4.5 by 1.5 gives a final value of 3.

Freezing point depression is a colligative property describing how the freezing point of a solvent decreases proportionally with the molality of a dissolved solute. The mass of the ethanol solvent is determined by multiplying its density of 0.80 g cm-3 by its volume of 62.5 cm3, yielding 50 g, which is 0.050 kg. The depression in the freezing point, calculated as the difference between the pure solvent and the solution, is 0.9 K. Using the relationship $\Delta T_f = K_f \cdot m$, the equation becomes $0.9 = 2.00 \times \frac{1.8 / M_A}{0.050}$, and rearranging to solve for the molar mass $M_A$ yields $M_A = \frac{2.00 \times 1.8}{0.9 \times 0.050}$, which equals 80 g mol-1.

The phenomenon of boiling point elevation occurs because non-volatile solute particles impede the transition of solvent molecules into the vapor phase, which necessitates a higher temperature for the system to reach its boiling point. This elevation, denoted as $\Delta T_b$, is related to the molality of the solution by the expression $\Delta T_b = \frac{K_b \times W_{\text{solute}}}{M_{\text{solute}} \times W_{\text{solvent}}}$, where the masses are in kilograms.

Subtracting the boiling point of pure water from the boiling point of the solution yields $\Delta T_b = 373.535 - 373.15 = 0.385$ K. Rearranging the relationship to solve for the molar mass of the solute provides $M_{\text{solute}} = \frac{K_b \times W_{\text{solute}}}{\Delta T_b \times W_{\text{solvent}}}$. Inserting the provided constants results in $M_{\text{solute}} = \frac{0.52 \times 2.5 \times 10^{-3}}{0.385 \times 75 \times 10^{-3}}$ kg mol$^{-1}$. Converting this result to grams by multiplying by 1000 produces approximately 45.02 g mol$^{-1}$, which rounds to 45 g mol$^{-1}$.