Solutions – JEE Main ChemistryPractice Questions & PYQs

The elevation of boiling point is a colligative property that occurs when a non-volatile solute is dissolved in a solvent, causing the solution to boil at a higher temperature than the pure solvent. This relationship is quantified by the equation $\Delta T_b = K_b \cdot m$, where $\Delta T_b$ is the elevation in boiling point, $K_b$ is the ebullioscopic constant, and $m$ is the molality of the solution.

Begin by determining the boiling point elevation, which is $373.52 - 373 = 0.52$ K. Since molality is the amount of solute in moles divided by the mass of the solvent in kilograms, substitute the known values into the expression $0.52 = 0.52 \times \frac{2}{M \times 0.02}$. Solving this equation leads to $1 = \frac{100}{M}$, which confirms that the molar mass of the solute is 100 g/mol.

Freezing point depression is a colligative property directly proportional to the total number of solute particles, and for dissociating electrolytes, this effect is quantified by the van't Hoff factor $i$ where the observed depression relates to the theoretical value as $\Delta T_{f(\text{obs})} = i \cdot \Delta T_{f(\text{calc})}$. Since a weak monobasic acid dissociates into two ions per molecule, the factor is calculated as $i = 1 + 0.3(2 - 1) = 1.3$. Because the observed freezing point depression scales by this factor relative to the theoretical value, the percentage increase is determined by the formula $\frac{i - 1}{1} \times 100$. Inserting the value of 1.3 into this expression, the calculation $(1.3 - 1) \times 100$ results in a 30 percent increase.

The van't Hoff factor, $i$, quantifies the deviation of observed colligative properties from ideal behavior and is mathematically defined as the ratio $\frac{\text{Normal molar mass}}{\text{Abnormal molar mass}}$. In the study of freezing point depression, the proportionality constant characteristic of a solvent is known as the cryoscopic constant, $k_f$. Systems that share an identical osmotic pressure are classified as isotonic, whereas azeotropes are defined as mixtures that boil at a constant temperature while maintaining the same composition in both the liquid phase and the vapor phase above it.

Osmotic pressure is a colligative property determined by the molar concentration of solute particles, described by the relation $\pi = CRT$. Since the osmotic pressure of the non-electrolyte solution A matches that of the glucose solution at the same temperature, their molar concentrations must be equivalent. The glucose solution has a concentration of 0.05 M.

Concentration is defined as the number of moles of solute divided by the volume of solution in liters. By dissolving 12 g of solute in 1000 mL, which is 1 L, the molarity becomes $\frac{12}{M_A}$ where $M_A$ is the molecular mass. Equating this to the concentration of the glucose solution results in the equation $\frac{12}{M_A} = 0.05$. Calculation of $M_A$ via $M_A = \frac{12}{0.05}$ yields a molecular mass of 240 g/mol.

Titration relies on stoichiometry to determine the amount of substance present in a solution. Because KCl reacts with AgNO3 in a 1:1 molar ratio, the 20 mL of 1 M AgNO3 used in the titration indicates that 0.02 moles of KCl were present in the initial sample. Determining the freezing point depression requires calculating the molality of this solution, which necessitates isolating the mass of the solvent. With a total volume of 25 mL and a density of 1 g/mL, the total solution mass is 25 g. Removing the mass of the KCl solute, which is calculated as 0.02 moles multiplied by 74.5 g/mol to give 1.49 g, leaves 23.51 g of solvent, or 0.02351 kg.

Dividing the number of moles by the mass of the solvent in kilograms gives a molality $m \approx 0.85$ mol/kg. Since KCl is an electrolyte that ionizes completely into two particles per formula unit, the van't Hoff factor is 2. Using the colligative property formula $\Delta T_f = i \cdot K_f \cdot m$, where $K_f = 2.0$ K kg/mol, the expression becomes $\Delta T_f = 2 \times 2.0 \times 0.85 = 3.4$ K. Rounding to the nearest integer, the final value is 3.

Osmotic pressure is determined by the van 't Hoff equation $\pi = CRT$, where $C$ represents the molar concentration of the solute. Expressing concentration as the ratio of mass $w$ to the product of molar mass $M$ and volume $V$, the relationship becomes $\pi = \frac{wRT}{MV}$. Using the provided gas constant of 0.083 L bar K$^{-1}$ mol$^{-1}$, the osmotic pressure of 400 Pa must be converted to 0.004 bar by dividing by 10^5, and the temperature is 300 K. Substituting the values w = 2.5 g and V = 0.250 L into the rearranged equation yields the expression

$M = \frac{2.5 \times 0.083 \times 300}{0.004 \times 0.250}$

which results in 62250 g/mol.

Osmotic pressure provides a method to determine the molar mass of proteins by relating the pressure exerted by the solute to the concentration of the solution. The relationship is governed by the equation
$\pi = \frac{wRT}{MV}$
where $\pi$ is the osmotic pressure, $w$ is the mass of the solute, $R$ is the universal gas constant, $T$ is the temperature, $M$ is the molar mass, and $V$ is the volume of the solution.

Converting the volume of $300 \text{cm}^3$ to $0.3 \text{L}$ and the osmotic pressure of $1.29 \text{mbar}$ to $1.29 \times 10^{-3} \text{bar}$ ensures that the units are consistent with the provided gas constant $R = 0.083 \text{L bar K}^{-1} \text{mol}^{-1}$. Rearranging the formula to isolate the molar mass gives $M = \frac{wRT}{\pi V}$. Substituting the known values $w = 0.63 \text{g}$, $T = 300 \text{K}$, $V = 0.3 \text{L}$, and $\pi = 1.29 \times 10^{-3} \text{bar}$ results in the calculation $M = \frac{0.63 \times 0.083 \times 300}{1.29 \times 10^{-3} \times 0.3}$, which evaluates to $40535 \text{g mol}^{-1}$.

The boiling point elevation constant, $K_b$, is determined by the relationship $K_b = \frac{R T_b^2 M}{1000 \Delta H_{vap}}$, where $T_b$ represents the boiling point, $M$ is the molar mass of the solvent, and $\Delta H_{vap}$ is the enthalpy of vaporization. For solvents $X$ and $Y$ with identical molecular weights, the term $M$ cancels out when comparing their respective constants, leaving the ratio of the elevation constants to depend entirely on the square of the boiling point ratio and the inverse of the enthalpy of vaporization ratio. Given the boiling point ratio of 2:1 and the enthalpy of vaporization ratio of 1:2, the calculation is:

$\frac{(K_b)_x}{(K_b)_y} = \left( \frac{2}{1} \right)^2 \times \left( \frac{2}{1} \right) = 4 \times 2 = 8$

The value of $m$ is 8.

Osmotic pressure is determined by the relationship
$\pi = iCRT$
where $i$ represents the van't Hoff factor, $C$ is the molar concentration, $R$ is the ideal gas constant, and $T$ is the temperature in Kelvin. Since $R$ and $T$ are constant, two solutions exhibit identical osmotic pressure if their $i \times C$ products are equal, assuming complete ionization of the solutes.

For the first pair, $0.500 \text{ M}$ ethanol ($i=1$) and $0.25 \text{ M}$ $\text{KBr}$ ($i=2$) both result in a product of $0.50$. In the second pair, $0.100 \text{ M}$ $\text{K}_4[\text{Fe}(\text{CN})_6]$ ($i=5$) and $0.100 \text{ M}$ Mohr's salt ($i=5$) yield an identical product of $0.50$. The third pair produces values of $0.25$ and $0.50$, indicating they are unequal. The fourth pair, $0.15 \text{ M}$ $\text{NaCl}$ ($i=2$) and $0.1 \text{ M}$ $\text{BaCl}_2$ ($i=3$), both result in a value of $0.3$. Finally, the fifth pair, carnallite ($i=5$ at $0.02 \text{ M}$) and $\text{KCl}$ ($i=2$ at $0.05 \text{ M}$), both yield $0.1$. Because four of these pairs demonstrate equivalent osmotic pressures, the total number of such pairs is $4.00$.

The presence of a non-volatile solute reduces the vapour pressure of a solvent, a phenomenon described by the relative lowering of vapour pressure. Because $\mathrm{MgCl}_2$ is an electrolyte, it partially dissociates in aqueous solution, increasing the total number of particles and necessitating the use of the van't Hoff factor to account for this increase. The dissociation reaction produces one magnesium ion and two chloride ions, meaning three ions form per formula unit. With a dissociation degree of 0.8, the van't Hoff factor is calculated as $i = 1 + (3 - 1) \times 0.8$, which equals 2.6.

A 1.0 molal solution contains 1 mole of solute dissolved in 1 kilogram of water. Since the molar mass of water is 18 g/mol, the solvent consists of $n_1 = \frac{1000}{18} \approx 55.55$ moles. The relationship for relative lowering of vapour pressure, $\frac{P^\circ - P_s}{P^\circ} = \frac{i \cdot n_2}{n_1}$, determines the pressure drop as $\Delta P = 50 \times \frac{2.6 \times 1}{55.55}$, which evaluates to approximately 2.34 mmHg. Subtracting this value from the initial vapour pressure of 50 mmHg results in a final vapour pressure of $50 - 2.34 = 47.66$ mmHg, which becomes 48 mmHg when rounded to the nearest integer.

Adding a non-volatile solute to a solvent blocks surface sites, hindering the escape of solvent molecules into the gas phase and inevitably causing the vapour pressure of the resulting solution to be lower than that of the pure solvent. During the process of freezing, the system reaches equilibrium where the vapour pressure of the liquid equals that of the solid, and at this stage, only the solvent molecules possess the affinity to transition into the organized solid crystal lattice, leaving the solute behind in the liquid.

A liquid reaches its boiling point when its vapour pressure increases to equal the standard atmospheric pressure. By identifying the intersection point on the vapour pressure versus temperature curve where this condition is met for the pure solvent, we can directly read the temperature from the horizontal axis. Following this relationship on the graph, the boiling point of the solvent is 82°C.

The boiling point of a solution is higher than that of the pure solvent because dissolved solute particles decrease the solvent's vapor pressure. For strong electrolytes, the magnitude of this boiling point elevation is proportional to the total concentration of ions, which is determined by multiplying the solute's molarity by its van 't Hoff factor, reflecting the number of ions produced per formula unit upon complete dissociation. Since sea water contains sodium chloride and magnesium chloride, the total boiling point elevation is calculated by summing the contributions of all resulting ions.

In 100 grams of the sea water, the mass of the water solvent is determined by the relation $100 - 29.25 - 19 = 51.75$ grams. Using the provided molar masses, the amount of sodium chloride is $29.25 / 58.5 = 0.5$ moles and the amount of magnesium chloride is $19 / 95 = 0.2$ moles. Accounting for complete ionization, where sodium chloride yields two ions and magnesium chloride yields three, the total moles of particles are calculated as $(2 \times 0.5) + (3 \times 0.2) = 1.6$ moles. The boiling point elevation $\Delta T_b$ is then found using the molality of these particles in the solvent:

$\Delta T_b = 0.52 \times \frac{1.6}{0.05175} \approx 16.07$

Adding this elevation to the boiling point of pure water results in $100 + 16.07 = 116.07$ degrees Celsius, which rounds to a nearest integer value of 116 degrees Celsius.

The phenomenon where adding a solute lowers the freezing point of a solvent is known as freezing point depression, mathematically expressed as $\Delta T_f = i \cdot K_f \cdot m$. A 38% solution by weight implies that 38 g of sulfuric acid is dissolved in every 100 g of total solution, leaving 62 g of water as the solvent. The molality is determined by dividing the moles of solute, 38 divided by 98, by the solvent mass of 0.062 kg. Using the van't Hoff factor of 2.67 and the cryoscopic constant of 1.8 K kg mol⁻¹, the total depression in freezing point is calculated as $\Delta T_f = 2.67 \times 1.8 \times \frac{38}{98} \times \frac{1000}{62}$. This results in a depression of approximately 30.05 K, and subtracting this value from the standard freezing point of pure water at 273 K determines the final temperature of 243 K.

When a solute like acetic acid undergoes partial dissociation, the freezing point depression depends on the van 't Hoff factor $i = 1 + \alpha(n-1)$ to account for the increased particle count. With a degree of dissociation $\alpha = 0.2$ and $n=2$ ions produced per molecule, the factor is calculated as $i = 1 + 0.2(2-1) = 1.2$. The molality is determined by dividing the $5/60$ moles of acetic acid by the $0.5$ kg mass of the solvent, resulting in $m = 1/6$ mol/kg. Substituting these values into the formula $\Delta T_f = i \times K_f \times m$ with the cryoscopic constant $K_f = 1.86$ K kg/mol leads to $\Delta T_f = 1.2 \times 1.86 \times (1/6) = 0.372$ °C, which equals $372 \times 10^{-3}$ °C.

The reduction in vapour pressure upon dissolving a non-volatile solute is governed by the principle of relative lowering of vapour pressure, where the ratio of the lowering to the solution's vapour pressure equals the ratio of the number of moles of solute to that of the solvent. Since a 25% reduction in vapour pressure means the final vapour pressure is 75% of the original, the ratio $\frac{P^{0-}P_s}{P_s}$ becomes $\frac{P^0 - 0.75 P^0}{0.75 P^0} = \frac{0.25 P^0}{0.75 P^0}$, which simplifies to $\frac{1}{3}$.

Calculating the required mass $x$ of urea begins with determining the moles of solvent by dividing the 1000 g of water by its molar mass of 18 g/mol, resulting in $\frac{1000}{18}$ moles. Given the molar mass of urea, $\mathrm{NH}_2\mathrm{CONH}_2$, is 60 g/mol, the number of moles of solute is $\frac{x}{60}$. Applying the relationship $\frac{n_{\text{solute}}}{n_{\text{solvent}}} = \frac{1}{3}$ yields the equation $\frac{x/60}{1000/18} = \frac{1}{3}$. Rearranging this equation to solve for $x$ gives $x = \frac{1000 \times 60}{3 \times 18} = \frac{10000}{9}$, which is approximately 1111 g.

Raoult's law describes the total vapor pressure of an ideal binary solution as the sum of partial pressures proportional to each component's mole fraction, following the relationship $P_{total} = X_A P_A^0 + X_B P_B^0$. Experimental observations lead to the equations $0.7 P_A^0 + 0.3 P_B^0 = 350$ and $0.2 P_A^0 + 0.8 P_B^0 = 410$. Multiplying the first equation by 8 yields $5.6 P_A^0 + 2.4 P_B^0 = 2800$, while multiplying the second by 3 results in $0.6 P_A^0 + 2.4 P_B^0 = 1230$. Subtracting the second equation from the first eliminates the contribution of component $B$, leaving $5.0 P_A^0 = 1570$, which identifies the vapor pressure of pure $A$ as 314 mm Hg.

Osmotic pressure serves as a colligative property that relates directly to the molar mass of a polymer. By expressing the ideal gas law relationship for osmotic pressure as $\pi = \frac{w}{MV} RT$, we can define mass concentration as $c = \frac{w}{V}$. This substitution allows us to write the relationship as $\pi = \left( \frac{RT}{M} \right) c$, indicating that a graph of osmotic pressure against mass concentration is linear with a slope equal to $\frac{RT}{M}$.

Using the provided slope of $6 \times 10^{-4}$, the gas constant $R = 0.083$ L atm K⁻¹ mol⁻¹, and temperature 300 K, we can define the relationship as $6 \times 10^{-4} = \frac{0.083 \times 300}{M}$ Rearranging this expression to solve for the molar mass $M$ results \in $M = \frac{0.083 \times 300}{6 \times 10^{-4}} = 41500$ g mol⁻¹.

The relative lowering of vapour pressure in a dilute solution corresponds to the mole fraction of the solute, which is defined by the relationship $\frac{P^0 - P_s}{P^0} = \frac{n}{N}$. With a pure water vapour pressure of 54.2 mmHg and a pressure reduction of 0.20 mmHg, the relative lowering ratio is $\frac{0.20}{54.2}$. The moles of the solvent are calculated using the 100 g sample of water and its molar mass of 18 g/mol, while the moles of solute are expressed as $\frac{w}{180}$ where $w$ represents the mass of glucose in grams. The equation $\frac{0.20}{54.2} = \frac{w/180}{100/18}$ results in a value of 3.69 g for the mass of glucose.

Linking colligative properties to the concentration of solute particles determines the initial amount of lead nitrate. The boiling point elevation $\Delta T_b = 0.15$ °C results from the dissociation of lead nitrate into one $\text{Pb}^{2+}$ ion and two $\text{NO}_3^{-}$ ions, establishing a van't Hoff factor $i = 3$. Assuming molality equals molarity, the expression $\Delta T_b = i K_b m$ becomes $0.15 = 3 \times 0.5 \times a$, where $a$ denotes the initial molarity of $\text{Pb(NO}_3)_2$. Solving this equation gives $a = 0.1$ mol, signifying the presence of $0.1$ mol of $\text{Pb}^{2+}$ and $0.2$ mol of $\text{NO}_3^{-}$.

Adding $0.2$ mol of $\text{NaCl}$ introduces $0.2$ mol each of $\text{Na}^{+}$ and $\text{Cl}^{-}$ ions, which initiates the precipitation reaction $\text{Pb}^{2+} + 2 \text{Cl}^{-} \rightarrow \text{PbCl}_2(\text{s})$. With $x$ representing the moles of $\text{PbCl}_2$ precipitated, the solution retains $(0.1 - x)$ mol of $\text{Pb}^{2+}$ and $(0.2 - 2x)$ mol of $\text{Cl}^{-}$. The freezing point depression $\Delta T_f = 0.8$ °C reflects the total ion concentration, which is the sum of the remaining $\text{Pb}^{2+}$, the $\text{Cl}^{-}$, the spectator $\text{NO}_3^{-}$ ($0.2$ mol), and the spectator $\text{Na}^{+}$ ($0.2$ mol). Applying $\Delta T_f = K_f \times (\text{total moles of ions})$ leads to $0.8 = 1.8 \times ((0.1 - x) + (0.2 - 2x) + 0.2 + 0.2)$. Simplifying this equation gives $0.8 = 1.8 \times (0.7 - 3x)$, which determines $x = \frac{2.3}{27}$ mol.

Calculating the equilibrium concentrations of the ions for the solubility product K_{sp} = \[\text{Pb}^{2+}]$$[\text{Cl}^{-}]^2$involves substituting$x$into the ionic concentrations. The concentration of$\text{Pb}^{2+}$is$0.1 - \frac{2.3}{27} = \frac{0.4}{27}$M, and the concentration of$\text{Cl}^{-}$is$0.2 - 2 × \frac{2.3}{27} = \frac{0.8}{27}$M. The solubility product is thus$K_{sp} = (\frac{0.4}{27}) × (\frac{0.8}{27})^2$, which yields$K_{sp} = \frac{0.4 × 0.64}{19683} \approx 13 × 10^{-6}$.