Solutions – JEE Main ChemistryPractice Questions & PYQs

The Van't Hoff factor quantifies the actual particle concentration in a solution relative to the theoretical concentration, directly influencing colligative properties like freezing point depression according to the equation $\Delta T_f = i K_f m$. To find this value, determine the mass of the formic acid solute by multiplying its volume concentration of $0.5 \text{ mL L}^{-1}$ by its density of $1.05 \text{ g mL}^{-1}$, which yields $0.525 \text{ g}$. Since the molar mass of formic acid is $46 \text{ g mol}^{-1}$, this mass corresponds to $\frac{0.525}{46}$ moles. Assuming the solution volume is based on $1 \text{ kg}$ of water, the molality is $\frac{0.525}{46 \times 1} \text{ mol kg}^{-1}$. Substituting the observed freezing point depression of $0.0405^{\circ} \text{C}$ and the cryoscopic constant of $1.86 \text{ K kg mol}^{-1}$ into the standard relationship creates the following expression:

$0.0405 = i \times 1.86 \times \frac{0.525}{46}$

Solving for $i$ by rearranging the terms results in a calculation of $\frac{0.0405 \times 46}{1.86 \times 0.525}$, which gives a value of approximately $1.9$.

Determining the vapor pressure of a pure component within an equilibrium mixture requires bridging gas-phase behavior with liquid-phase composition using Dalton's Law and Raoult's Law. Dalton's Law indicates that the partial pressure of a specific gas in a mixture is the product of its mole fraction in the vapor phase and the total pressure, expressed as $P_A = Y_A \times P_T$. Inserting the provided values, we calculate $P_A = 0.5 \times 0.8 = 0.4$ atm. According to Raoult's Law, this partial pressure is also related to the liquid-phase mole fraction and the vapor pressure of the pure substance via the equation $P_A = X_A \times P_A^\circ$. Substituting the known values into $0.4 = 0.2 \times P_A^\circ$ allows us to solve for $P_A^\circ = 0.4 / 0.2$, which results in 2 atm.

Osmotic pressure acts as a colligative property that depends on the concentration of solute particles, allowing for the determination of the molar mass of complex molecules like proteins using the equation $\pi = \left(\frac{m}{M \cdot V}\right) RT$, where $\pi$ is osmotic pressure, $m$ is the solute mass, $M$ is molar mass, $V$ is volume in liters, $R$ is the gas constant, and $T$ is temperature. Substituting the provided values into this expression gives $5.03 \times 10^{-3} = \frac{2.5}{M} \times \left(\frac{1000}{500}\right) \times 0.083 \times 300$, which results in a molar mass of $24751.5\ \mathrm{g/mol}$ for the protein. Given that glycine ($\mathrm{C}_2\ \mathrm{H}_5\ \mathrm{NO}_2$) has a molar mass of $75\ \mathrm{g/mol}$, dividing the protein's total molar mass by this value yields a total of 330 glycine units.

The depression in freezing point serves as a colligative property proportional to the concentration of solute particles, which is expressed by the relationship $\Delta T_f = K_f \times \frac{w}{M \times W}$. In this scenario, the mass of the solutes $w$ and the mass of the solvent $W$ remain constant for both solutions, making the freezing point depression inversely proportional to the molar mass $M$ of the solute.

Since the ratio of the freezing point depressions is $\frac{\Delta T_{f(A)}}{\Delta T_{f(B)}} = \frac{1}{4}$, the molar masses follow the inverse ratio where $\frac{M_Y}{M_X} = \frac{1}{4}$. This equality indicates that $M_X = 4 M_Y$, meaning the molar mass of $X$ is four times larger than the molar mass of $Y$. Expressing the ratio of $M_X : M_Y$ as $4 : 1$ corresponds directly to the value $1 : 0.25$.

Henry's law states that the partial pressure of a gas is directly proportional to its mole fraction in a solution, expressed as $P = K_H X$. With a partial pressure of 0.835 bar and a Henry's law constant of 1.67 kbar, the mole fraction of carbon dioxide is:

$X = \frac{0.835}{1670} = 5 \times 10^{-4}$

The total moles in one litre of water, which weighs 1000 g, is approximately 55.5 mol. Multiplying the mole fraction by this value gives the moles of dissolved carbon dioxide as 0.02775 mol. Using the molar mass of carbon dioxide, 44 g mol$^{-1}$, the final mass calculation results in 1.221 g, which matches the required value of $1221 \times 10^{-3}$ g.

Freezing point depression is a colligative property that depends on the concentration of solute particles in a solution, expressed by the equation $\Delta T_f = K_f \cdot m$. This principle dictates that the decrease in the freezing point is directly proportional to the molality of the solute, given the constant $K_f$.

Calculating the molality of the ascorbic acid solution involves dividing the 10.2 g of solute by its molar mass of 176 g/mol to find the moles of solute, and then dividing that amount by the solvent mass of 0.15 kg. Substituting these values into the freezing point depression equation provides the magnitude of the freezing point change:
$\Delta T_f = 3.9 \times \frac{10.2 / 176}{0.15}$

Solving this expression results in $\Delta T_f \approx 1.506$. Expressing this value in the requested form of $15.06 \times 10^{-1}$ allows us to identify the integer value of $x$ as 15.

Freezing point depression is a colligative property determined by the total concentration of particles in a solution, meaning the van't Hoff factor must be included to account for solute dissociation. Since chloroacetic acid dissociates according to the equilibrium $\mathrm{ClCH}_2\mathrm{COOH} \rightleftharpoons \mathrm{ClCH}_2\mathrm{COO}^{-} + \mathrm{H}^{+}$, the total number of particles in the solution is $1 + \alpha$, where $\alpha$ represents the degree of dissociation. To find the concentration, we determine the molality $m$ by dividing the moles of solute, $\frac{9.45\,\mathrm{g}}{94.5\,\mathrm{g/mol}} = 0.1\,\mathrm{mol}$, by the solvent mass of $0.5\,\mathrm{kg}$, resulting in $0.2\,\mathrm{mol/kg}$. Using the freezing point depression formula $\Delta T_{f} = i K_{f} m$, we substitute the known values to obtain $0.5 = (1 + \alpha) \times 1.86 \times 0.2$. Simplifying the right side results in $0.5 = (1 + \alpha) \times 0.372$, and solving for $1 + \alpha$ gives $\frac{0.5}{0.372} \approx 1.344$. Subtracting $1$ from this value yields $\alpha = 0.344$, which corresponds to a percentage of $34.4\%$. Consequently, the value of $x$ is $34.4$.

Freezing point depression occurs because adding a solute disrupts the crystallization process of the solvent, resulting in a temperature drop defined by the equation $\Delta T_f = i K_f m$. The molar mass of the butane solute is 58 grams per mole, meaning 10 grams represents approximately 0.172 moles. Dividing this by the 0.2 kilograms of benzene yields a molality of 0.862 molal. With the given constant of 5.12 degrees Celsius per molal and a van't Hoff factor of 1, the depression equals 5.12 multiplied by 0.862, which is approximately 4.41 degrees Celsius. Subtracting this value from the original 5.5 degrees Celsius results in a final freezing point of 1 degree Celsius.

Colligative properties depend on the total number of solute particles in a solution, meaning that when a solute undergoes dimerization, the effective particle count decreases. To account for this deviation from ideal behavior, we use the van't Hoff factor, denoted as $i$, which relates the observed colligative effect to the theoretical concentration.

The elevation in boiling point is determined by the difference between the solution temperature and that of pure water, resulting in $\Delta T_b = 0.52$ K. Substituting this into the formula $\Delta T_b = i \times K_b \times m$ gives $0.52 = i \times 0.52 \times 2$, which simplifies to $i = 0.5$. Because the solute dimerizes, the relationship involving the degree of association $\alpha$ is $i = 1 - \frac{\alpha}{2}$. Solving $0.5 = 1 - \frac{\alpha}{2}$ results in $\alpha = 1$, which corresponds to an association of 100 percent.

The freezing point of a solution decreases when a non-volatile solute is added, a change quantified by the relationship $\Delta T_f = K_f m$. First, determine the amount of glucose in moles by calculating $\frac{40}{180} = 0.22$ mol. With a water density of 1.00 g cm$^{-3}$, the 200 mL of solvent is equivalent to 200 g, which is 0.2 kg. The molality $m$ is then determined by the expression $\frac{0.22}{0.2} = 1.1$ mol kg$^{-1}$. Substituting the given cryoscopic constant into the depression formula gives $\Delta T_f = 1.86 \times 1.1 = 2.046$ K, which is a decrease of approximately 2 K. Subtracting this from the freezing point of pure water of 273.15 K results in 271.15 K, which rounds to 271 K.

Boiling point elevation is a colligative property that depends on the total concentration of particles present in a solution, requiring an adjustment for solute dissociation using the van't Hoff factor. Since $A B_{2}$ dissociates into three ions, specifically one $A^{2+}$ and two $B^{-}$ ions, the total number of ions per formula unit is $n=3$. With a degree of dissociation of $10\%$, or $\alpha = 0.1$, the van't Hoff factor is calculated as $i = 1 + (n-1)\alpha$, which results in $i = 1 + (3-1)(0.1) = 1.2$.

Applying this factor to the boiling point elevation formula, $\Delta T_{b} = i K_{b} m$, we substitute the known values where $i = 1.2$, the molal elevation constant is $K_{b} = 0.5\,\mathrm{K}\,\mathrm{kg}\,\mathrm{mol}^{-1}$, and the molality is $m = 10.0$. This calculation yields $\Delta T_{b} = 1.2 \times 0.5 \times 10 = 6^{\circ}\mathrm{C}$. Adding this elevation to the standard boiling point of pure water results in a final boiling point of $100^{\circ}\mathrm{C} + 6^{\circ}\mathrm{C} = 106^{\circ}\mathrm{C}$.

Raoult's Law provides the foundation for finding the total vapour pressure of an ideal mixture by summing the individual partial pressures of its components. Each component's partial pressure is determined by multiplying its vapour pressure in a pure state by its specific mole fraction within the solution.

With one mole of A and two moles of B, the total composition comprises three moles. This results in mole fractions of $\chi_A = \frac{1}{3}$ and $\chi_B = \frac{2}{3}$. Using the relation $P_T = P_A^\circ \chi_A + P_B^\circ \chi_B$, the calculation involves multiplying 21 by 1/3 and 18 by 2/3. This gives a partial pressure of 7 kPa for A and 12 kPa for B, leading to a total vapour pressure of 19 kPa.

Freezing point depression is a colligative property governed by the equation $\Delta T_f = i K_f m$, which dictates that for a fixed solute concentration, the freezing point decreases more significantly as the van't Hoff factor $i$ increases. Among the choices, the solute $\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6$ has $i$ equal to 1, $\mathrm{KI}$ has $i$ equal to 2, $\mathrm{K}_2\mathrm{SO}_4$ has $i$ equal to 3, and $\mathrm{Al}_2(\mathrm{SO}_4)_3$ has $i$ equal to 5. Since the $\mathrm{Al}_2(\mathrm{SO}_4)_3$ solution generates the highest number of particles per mole, it experiences the maximum depression, leading to the lowest freezing point.

The boiling point elevation of a solution is a colligative property governed by the equation $\Delta T_{b} = K_{b} \times m$, where $\Delta T_{b}$ is the change in boiling point, $K_{b}$ is the ebullioscopic constant of the solvent, and $m$ represents the molality.

Determining the molar mass $M$ requires expressing molality as the moles of solute, $\frac{3}{M}$, divided by the solvent mass of 0.100 kg. Applying the given values yields $0.60 = 5.0 \times \frac{3}{0.100M}$. This simplifies to $0.60 = \frac{150}{M}$, which rearranges to $M = \frac{150}{0.60}$, resulting in a final molar mass of 250 g/mol.

To determine the lowering of vapor pressure, one must first identify the amount of non-volatile solute formed through the chemical reaction between sulfur dioxide and sodium hydroxide. By applying the ideal gas law, $n = \frac{PV}{RT}$, the number of moles of sulfur dioxide is $\frac{1 \times 0.224}{0.082 \times 298}$, which equals $0.00916 \text{ mol}$. Given that $100 \text{ mL}$ of $0.1 \text{ M} \mathrm{NaOH}$ solution contains $0.01 \text{ mol}$ of base, the balanced reaction $\mathrm{SO}_2 + 2\mathrm{NaOH} \rightarrow \mathrm{Na}_2\mathrm{SO}_3 + \mathrm{H}_2\mathrm{O}$ shows that the reaction is limited by the amount of $\mathrm{NaOH}$. Because two moles of $\mathrm{NaOH}$ are required for every mole of $\mathrm{SO}_2$, the amount of $\mathrm{Na}_2\mathrm{SO}_3$ produced is exactly half the amount of $\mathrm{NaOH}$ present, resulting in $0.005 \text{ mol}$ of solute.

Once the solute is introduced into $36 \text{ g}$ of water, which corresponds to $\frac{36}{18} = 2 \text{ mol}$ of solvent, one can apply the principles of colligative properties. Since $\mathrm{Na}_2\mathrm{SO}_3$ is a strong electrolyte that dissociates into three ions-two $\mathrm{Na}^{+}$ cations and one $\mathrm{SO}_3^{2-}$ anion-the van't Hoff factor is $i = 3$. According to Raoult's law for the relative lowering of vapor pressure, the change in pressure is given by $\Delta p = p^{\circ} \times i \times \frac{n_{\text{solute}}}{n_{\text{solvent}} + n_{\text{solute}}}$. Using $24 \text{ mm Hg}$ for $p^{\circ}$, this calculation becomes $\Delta p = 24 \times 3 \times \frac{0.005}{2 + 0.005}$, which equals approximately $0.1795 \text{ mm Hg}$. Rounding this to $0.18 \text{ mm Hg}$ confirms that $x \times 10^{-2} = 18 \times 10^{-2}$, establishing the value of $x$ as $18$.

Freezing point depression is a colligative property that depends on the total concentration of solute particles in a solution. When benzoic acid molecules undergo complete association, the number of particles effectively decreases, which is accounted for by the van 't Hoff factor, $i$. For the association of $n$ molecules into a single aggregate, the relationship is given by the formula $i = 1/n$.

The molality of the solution is determined by dividing the moles of benzoic acid by the mass of the solvent in kilograms. Given 12.2 g of benzoic acid with a molar mass of 122 g mol^-1, the amount of solute is 0.1 mol, which, dissolved in 0.1 kg of water, results in a molality of 1 mol kg^-1. Using the expression $\Delta T_f = i \cdot K_f \cdot m$, we substitute the observed freezing point depression of 0.93 K and the cryoscopic constant of 1.86 K kg mol^-1. This leads to the equation $0.93 = i \times 1.86 \times 1$. Solving for $i$ yields 0.5, and equating this to $1/n$ results in $0.5 = 1/n$, confirming that the number of associated benzoic acid molecules is $n = 2$.

When electrolytes dissolve, they partially dissociate into multiple ions, increasing the total concentration of particles and thereby raising the boiling point of the solvent. We determine the Van't Hoff factor $i$ using the dissociation reaction $A_{2} B_{3} \rightarrow 2A^{+3} + 3B^{-2}$, where the total number of ions produced per formula unit $n = 5$. Since the solute is 60% ionized, the degree of dissociation $\alpha = 0.6$, giving the formula $i = 1 + (n - 1)\alpha$, which calculates to $i = 1 + (5 - 1) \times 0.6 = 3.4$.

With the Van't Hoff factor determined, we apply the boiling point elevation formula $\Delta T_b = i \cdot K_b \cdot m$ to find the temperature increase. Substituting the given values yields $\Delta T_b = 3.4 \times 0.52 \times 1 = 1.768$ K. The final boiling point is then found by adding this elevation to the standard boiling point of water, $T_b = 373.15 + 1.768 = 374.918$ K. Rounding this value to the nearest integer gives 375 K.

The composition of the vapour phase above a liquid mixture is dictated by Dalton's law of partial pressures, where the mole fraction of a component in the vapour is the ratio of its partial pressure to the total pressure. For an equimolar mixture, the liquid mole fraction of both benzene and methyl benzene is $0.5$, resulting in partial pressures of $p_B = 0.5 \times 70 = 35$ torr and $p_T = 0.5 \times 20 = 10$ torr. Summing these gives a total pressure of $45$ torr, which makes the mole fraction of benzene in the vapour phase $y_B = \frac{35}{45} \approx 0.777$. Rounding this result to the nearest integer gives $78 \times 10^{-2}$.

Freezing point depression is a colligative property defined by the relationship $\Delta T_{f} = i \cdot K_{f} \cdot m$, which indicates that the extent of freezing point lowering is directly proportional to the number of solute particles present in the solution. Since each option has an identical concentration of 0.10 M, the variation in freezing point depression depends entirely on the van't Hoff factor ($i$), which represents the number of particles produced by each solute unit. Hydrazine, glucose, and glycine are non-electrolytes that remain as single molecules, whereas $\mathrm{KHSO}_{4}$ dissociates into $K^{+}$ and $HSO_{4}^{-}$ ions, resulting in a higher particle count and the largest freezing point depression.

The van't Hoff factor quantifies how association and dissociation processes alter the total number of particles relative to the amount of solute originally dissolved. When starting with $1$ mole of $\mathrm{HA}$, the conditions specified imply that $0.5$ moles undergo dimerization while the remaining $0.5$ moles undergo dissociation.

In the dimerization process, the stoichiometry $2\mathrm{HA} \rightarrow \mathrm{H}_2\mathrm{A}_2$ dictates that the $0.5$ moles of $\mathrm{HA}$ combine to form $0.25$ moles of the dimer. For the dissociation process, the reaction $\mathrm{HA} \rightarrow \mathrm{H}^+ + \mathrm{A}^-$ causes the $0.5$ moles of $\mathrm{HA}$ to break apart entirely, yielding $0.5$ moles of $\mathrm{H}^+$ and $0.5$ moles of $\mathrm{A}^-$. By adding these final amounts, the solution contains $0.25 + 0.5 + 0.5 = 1.25$ moles of particles. Dividing this total by the initial $1$ mole yields a van't Hoff factor of $1.25$, which is equivalent to $125 \times 10^{-2}$.