Solutions – JEE Main ChemistryPractice Questions & PYQs

According to Raoult's Law, the total vapour pressure of a mixture is the sum of the individual partial pressures exerted by each component, given by the relationship $P_{\text{total}} = P_A^{\circ} x_A + P_B^{\circ} x_B$. Because the sum of mole fractions must equal 1, the mole fraction of propyl alcohol is 0.4. The relationship for this mixture is:

$290 = P_A^{\circ} \times 0.6 + 200 \times 0.4$

The partial pressure contribution from propyl alcohol is 80 mm, so the partial pressure of ethyl alcohol must be 210 mm. Dividing this by the mole fraction 0.6 yields a pure vapour pressure of 350 mm for ethyl alcohol.

According to Raoult's law, the vapour pressure of an ideal solution containing a non-volatile solute is equal to the mole fraction of the solvent multiplied by the vapour pressure of the pure solvent. At 100°C, pure water exerts a vapour pressure of 760 Torr. To find the solution pressure, first calculate the moles of solute and solvent: glucose contributes $\frac{18}{180} = 0.1$ moles and water contributes $\frac{178.2}{18} = 9.9$ moles. The total number of moles present in the mixture is $0.1 + 9.9 = 10$, which makes the mole fraction of water $\frac{9.9}{10}$. Multiplying this mole fraction by the pure solvent pressure yields a final vapour pressure of $\frac{9.9}{10} \times 760 = 752.4$ Torr.

Dipole-dipole interactions occur exclusively between molecules that possess permanent net dipole moments. Because both acetonitrile ($\mathrm{CH}_3\mathrm{CN}$) and acetone are polar molecules with permanent charge separation, they interact primarily through these dipole-dipole forces. In the other scenarios, $\mathrm{KCl}$ and water engage in ion-dipole interactions, benzene and ethanol result in dipole-induced dipole attractions, and the nonpolar pairing of benzene and carbon tetrachloride ($\mathrm{CCl}_4$) relies on instantaneous dipole-induced dipole interactions.

Boiling point elevation and freezing point depression are classified as colligative properties, which means their magnitude is determined solely by the number of solute particles present rather than the chemical identity of the solute itself. Since an equimolar concentration implies an identical number of particles for any normal solute within the same solvent, the degree of change in temperature will be uniform across such solutions. Consequently, these solutions necessarily possess identical boiling and freezing points.

Raoult's Law dictates that the partial vapour pressure of a volatile component in an ideal solution is the product of its mole fraction and its pure vapour pressure, expressed as $P_B = X_B \times P_B^{\circ}$. First, determine the molar amounts by dividing the given masses by their molar masses, where 78 g of benzene yields 1 mole and 46 g of toluene yields 0.5 moles. The mole fraction of benzene is the ratio of its moles to the total amount of substance in the mixture, calculated as 1 divided by 1.5, or 2/3. Multiplying this mole fraction by the pure vapour pressure of benzene of 75 torr results in a partial vapour pressure of 50 torr.

The Van't Hoff factor quantifies the deviation of a solution's properties from those of an ideal non-electrolyte by representing the ratio of total particles at equilibrium to the initial number of solute moles. For the dissociation of $\mathrm{Na}_2\mathrm{SO}_4$, each formula unit breaks apart into two $\mathrm{Na}^+$ ions and one $\mathrm{SO}_4^{2-}$ ion. Starting with one mole of the salt, a degree of dissociation of $\alpha$ leaves $(1-\alpha)$ moles of undissociated salt, while producing $2\alpha$ moles of sodium ions and $\alpha$ moles of sulfate ions. Summing these individual components yields a total of $(1-\alpha) + 2\alpha + \alpha$ particles, which simplifies to $1+2\alpha$.

The relationship between molecular weight and gas density under the assumption of ideal behavior is defined by the equation $\rho = \frac{PM}{RT}$. Because ethanol ($\mathrm{C}_2\mathrm{H}_5\mathrm{OH}$) and dimethyl ether ($\mathrm{CH}_3\mathrm{OCH}_3$) are structural isomers that share the same molecular formula, $\mathrm{C}_2\mathrm{H}_6\mathrm{O}$, they possess identical molar masses ($M$), ensuring their gaseous densities are equal when measured at the same temperature ($T$) and pressure ($P$). Other parameters such as boiling point, heat of vaporization, and vapor pressure depend heavily on intermolecular forces, specifically the hydrogen bonding present in ethanol but absent in dimethyl ether, which causes these values to differ significantly.

The freezing point depression of a solution is governed by the expression $\Delta T_f = K_f \times m \times i$, where $K_f$ is the cryoscopic constant specific to the solvent. Because every solvent possesses a distinct constant, even solutions with identical molality prepared in different solvents will exhibit different freezing point depressions. Therefore, the statement claiming these solutions would result in the same freezing point depression is false.

Positive deviation from Raoult's law occurs when the attractive interactions between solute and solvent molecules are weaker than the cohesive forces present in the individual pure components, causing the molecules to escape into the vapor phase more readily. In a mixture of benzene and methanol, the non-polar benzene molecules distribute themselves between the methanol molecules, effectively disrupting the strong hydrogen bonding network characteristic of the pure alcohol. This weakening of the net intermolecular forces increases the overall escaping tendency of the components, resulting in a total vapor pressure higher than that predicted by ideal behavior, or $P_{total} > P_{ideal}$.

Elevation in boiling point is a colligative property directly proportional to the total concentration of solute particles present in a solution. Since glucose and urea are non-electrolytes, they do not dissociate and maintain effective particle concentrations of 0.015 M. In contrast, ionic compounds dissociate, meaning 0.01 M $\mathrm{KNO}_3$ splits into two ions per formula unit to yield an effective concentration of 0.02 M, whereas 0.01 M $\mathrm{Na}_2\mathrm{SO}_4$ dissociates into three ions per formula unit, resulting in the highest effective concentration of 0.03 M. Because the boiling point elevation scales with the total number of particles, the solution with the greatest particle concentration exhibits the highest boiling point.

An ideal solution is characterized by the absence of net energy changes upon mixing, as the intermolecular forces between distinct components are identical to those between identical molecules. Since the energy required to disrupt the existing intermolecular bonds is perfectly balanced by the energy released upon forming new A-B interactions, the enthalpy of mixing is $\Delta H_{\text{mix}} = 0$. Unlike the entropy and free energy of mixing, which are non-zero due to the inherent increase in randomness during the mixing process, the enthalpy specifically remains unchanged.

Freezing point depression is a colligative property that depends on the total number of particles in a solution, which requires calculating the Van't Hoff factor $i$ for the weak acid $HX$. As the acid partially ionizes according to the equilibrium $HX \rightleftharpoons H^+ + X^-$, the Van't Hoff factor is determined by $i = 1 + \alpha$, which gives $i = 1 + 0.3 = 1.3$.

The magnitude of the freezing point depression is calculated using the formula $\Delta T_f = i \cdot K_f \cdot m$. Substituting the given values into the expression yields $\Delta T_f = 1.3 \times 1.85 \times 0.2 = 0.481^{\circ}\,\mathrm{C}$. Because pure water freezes at $0^{\circ}\,\mathrm{C}$, the freezing point of the solution is $0 - 0.481^{\circ}\,\mathrm{C} = -0.481^{\circ}\,\mathrm{C}$, a value nearest to $-0.480^{\circ}\,\mathrm{C}$.

The primary factor that makes a pressure cooker efficient is the direct relationship between external pressure and the boiling point of water, represented by the principle that as pressure increases, the boiling point $T_{bp}$ increases. By sealing the vessel, the internal pressure builds up, which forces the water to remain in a liquid state at temperatures well beyond the standard boiling point. This high-temperature environment facilitates faster heat transfer into the food, thereby reducing the overall cooking time.

Boiling point elevation $\Delta T_b$ and freezing point depression $\Delta T_f$ are both colligative properties that depend on the molality of the solute particles in the solution. Because the solution is identical in both cases, the molality remains constant, which allows us to relate the boiling point elevation and freezing point depression directly through their respective constants using the expression $\frac{\Delta T_b}{\Delta T_f} = \frac{K_b}{K_f}$.

With an aqueous freezing point of $-0.186^\circ \text{C}$, the value of $\Delta T_f$ is $0.186^\circ \text{C}$. Substituting the constants $K_b = 0.512$ and $K_f = 1.86$ into the ratio formula leads to $\Delta T_b = \frac{0.512}{1.86} \times 0.186$. Since $0.186$ divided by $1.86$ simplifies to $0.1$, the final boiling point elevation is $0.512 \times 0.1$, which equals $0.0512^\circ \text{C}$.

Negative deviation from Raoult's law arises when the intermolecular attractive forces between unlike molecules are stronger than the forces between like molecules. Since the A-B interaction is stronger than the individual A-A and B-B interactions, the mixture experiences a contraction and releases heat, which is characterized by $\Delta V_{\text{mix}} < 0$ and $\Delta H_{\text{mix}} < 0$.