Solutions – JEE Main ChemistryPractice Questions & PYQs

The relative lowering of vapour pressure for a solution containing an electrolyte solute is given by $\frac{P^\circ - P_s}{P^\circ} = i \cdot x_2$, and the elevation in boiling point is given by $\Delta T_b = i \cdot K_b \cdot m$, where $i$ is the Van't Hoff factor, $x_2$ is the mole fraction of the solute, $m$ is the molality, and $K_b$ is the ebullioscopic constant.

Calculating the relative lowering of vapour pressure using the given values yields $\frac{640 - 600}{640} = i \cdot x_2$, which simplifies to $i \cdot x_2 = \frac{40}{640} = \frac{1}{16}$. For the elevation in boiling point, using the mass of water as $0.1\,\mathrm{kg}$ (since $100\,\mathrm{mL}$ of water has a mass of $100\,\mathrm{g}$), the expression becomes $375 - 373 = i \cdot 0.52 \cdot \frac{W/M}{0.1}$. This equation simplifies to $2 = i \cdot 5.2 \cdot \frac{W}{M}$, allowing for the determination of the Van't Hoff factor as $i = \frac{2}{5.2} \cdot \frac{M}{W} = \frac{1}{2.6} \cdot \frac{M}{W}$.

Substituting this expression for $i$ into the equation for relative vapour pressure lowering gives $x_2 = \frac{1}{16i} = \frac{1}{16 \cdot (\frac{1}{2.6} \cdot \frac{M}{W})} = \frac{2.6}{16} \cdot \frac{W}{M}$. Reducing the constant fraction $\frac{2.6}{16}$ by dividing the numerator and denominator by 2 results in $\frac{1.3}{8} \cdot \frac{W}{M}$.

The mole fraction of a component in a mixture is expressed as $x_i = \frac{n_i}{n_{\text{total}}}$, calculated by dividing the moles of the specific component by the total moles present in the solution. The molar mass of $\mathrm{H}_2\mathrm{SO}_4$ is $98\text{ g/mol}$, and the molar mass of $\mathrm{H}_2\mathrm{O}$ is $18\text{ g/mol}$. In the first solution, the mass of $\mathrm{H}_2\mathrm{SO}_4$ is $98\text{ g}$, yielding $1\text{ mol}$, and the mass of $\mathrm{H}_2\mathrm{O}$ is $2\text{ g}$, yielding $\frac{2}{18}\text{ mol}$. In the second solution, the mass of $\mathrm{H}_2\mathrm{SO}_4$ is $49\text{ g}$, yielding $0.5\text{ mol}$, and the mass of $\mathrm{H}_2\mathrm{O}$ is $51\text{ g}$, yielding $\frac{51}{18}\text{ mol}$.

Combining these amounts results in a total of $1.5\text{ mol}$ of $\mathrm{H}_2\mathrm{SO}_4$ and $\frac{53}{18}\text{ mol}$ of $\mathrm{H}_2\mathrm{O}$. Substituting these values into the mole fraction formula yields $x = \frac{1.5}{1.5 + (53/18)}$, which simplifies to $\frac{1.5}{(27/18) + (53/18)} = \frac{1.5}{80/18} = \frac{27}{80} = 0.3375$.

Osmotic pressure manifests as the hydrostatic head resulting from solvent flow across a semi-permeable membrane, and its value is determined by the expression $\pi = \rho gh$. Using the provided density $\rho = 1000 \, \mathrm{kg} \, \mathrm{m}^{-3}$, acceleration due to gravity $g = 10 \, \mathrm{ms}^{-2}$, and a height $h = 0.08 \, \mathrm{m}$, the pressure is calculated as $1000 \times 10 \times 0.08 = 800 \, \mathrm{Pa}$, which is equivalent to $0.8 \, \mathrm{kPa}$.

The relationship between osmotic pressure, solute concentration, and temperature is defined by the van 't Hoff equation $\pi = CRT$, which can be expanded as $\pi = \frac{w}{MV} RT$ where $w$ is the mass of the solute, $M$ is the molar mass, and $V$ is the volume. Substituting $0.8 \, \mathrm{kPa}$ for $\pi$, $20 \, \mathrm{g}$ for $w$, $1 \, \mathrm{L}$ for $V$, $8.3 \, \mathrm{kPa} \, \mathrm{dm}^{3} \, \mathrm{K}^{-1} \, \mathrm{mol}^{-1}$ for $R$, and $300 \, \mathrm{K}$ for $T$, the equation becomes $0.8 = \frac{20}{M \times 1} \times 8.3 \times 300$. Solving for $M$ results in $M = \frac{20 \times 8.3 \times 300}{0.8}$, which equals $62250 \, \mathrm{g} \, \mathrm{mol}^{-1}$ or $62.25 \, \mathrm{kg} \, \mathrm{mol}^{-1}$. Rounded to the nearest integer, the molar mass is $62 \, \mathrm{kg} \, \mathrm{mol}^{-1}$.

The boiling point elevation of a solution is given by $\Delta T_b = i \times K_b \times m$, where $i$ is the van't Hoff factor, $K_b$ is the ebullioscopic constant, and $m$ is the molality. For aqueous solutions of the same solvent, the boiling point elevation is directly proportional to the product of the van't Hoff factor and the molarity ($i \times M$), as molality is approximately equal to molarity in dilute solutions.

For glucose (I), $i=1$ and the molarity is $M = \frac{2.2/180}{0.125} = 0.0976$, resulting in a value of $0.0976$. For calcium chloride (II), which dissociates as $CaCl_2 \rightarrow Ca^{2+} + 2Cl^-$, $i=3$ and the molarity is $M = \frac{1.9/111}{0.25} = 0.0684$, resulting in $3 \times 0.0684 = 0.2052$. For urea (III), $i=1$ and the molarity is $M = \frac{9.0/60}{0.5} = 0.3$, resulting in $0.3$. For aluminium sulphate (IV), which dissociates as $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$, $i=5$ and the molarity is $M = \frac{20.5/342}{0.75} = 0.0799$, resulting in $5 \times 0.0799 = 0.3995$.

Comparing the calculated values of $0.0976$, $0.2052$, $0.3$, and $0.3995$, the increasing order of boiling point is I $<$ II $<$ III $<$ IV.

Osmosis is the spontaneous migration of solvent molecules through a semipermeable membrane from a region of lower solute concentration to a region of higher solute concentration to achieve equilibrium. By calculating the concentration of the glucose solution in each chamber, we find that Chamber 1 has a molarity of $1\ \mathrm{M}$, derived from $0.1\ \mathrm{mol}$ of glucose in $0.1\ \mathrm{L}$ of solution, while Chamber 2 has a molarity of $2/3\ \mathrm{M}$, derived from approximately $0.167\ \mathrm{mol}$ of glucose in $0.25\ \mathrm{L}$ of solution. Because water naturally flows from the lower concentration in Chamber 2 to the higher concentration in Chamber 1, the claim that water moves from Chamber 1 to Chamber 2 is incorrect.

The osmotic pressure of an electrolytic solution is determined by the equation $\pi = iCRT$, where $i$ is the Van't Hoff factor. Since potassium sulphate, $\mathrm{K}_{2}\mathrm{SO}_{4}$, dissociates into three ions per formula unit, the value of $i$ is $3$. Applying the given parameters, where $C = \frac{50 \times 10^{-3}\ \mathrm{g}}{174\ \mathrm{g/mol} \times 2\ \mathrm{L}}$, $T = 300\ \mathrm{K}$, and $R = 0.083\ \mathrm{dm}^{3}\ \mathrm{bar}\ \mathrm{K}^{-1}\ \mathrm{mol}^{-1}$, the calculation is:

$\pi = 3 \times \left( \frac{50 \times 10^{-3}}{174 \times 2} \right) \times 0.083 \times 300 = 0.0107\ \mathrm{bar}$

This result confirms that the calculated osmotic pressure matches the value provided in the statement, making the second assertion true.

Raoult's Law establishes that for a non-volatile, non-ionizable solute, the ratio of the solution's vapor pressure to the pure solvent's vapor pressure is identical to the mole fraction of the solvent. Determining this mole fraction involves dividing the number of moles of solvent, which is 1, by the total number of moles in the mixture, calculated as the sum of solvent and solute moles, $1 + 0.25 = 1.25$. This calculation, $\chi_A = \frac{1}{1.25}$, results in a value of $0.8$. Converting this figure to a percentage confirms that the vapor pressure of the solution is $80\%$ of the vapor pressure of the pure solvent.

The mixture of acetone and $CS_2$ exhibits positive deviation from Raoult's law because the intermolecular attractions between unlike acetone and $CS_2$ molecules are weaker than those between pure component molecules, resulting in a higher vapor pressure than predicted. While ideal behavior follows the linear relationship $P_{CS_2} = P_{CS_2}^0 x_{CS_2}$, the positive deviation causes the actual partial pressure of $CS_2$ to remain higher than the theoretical value at every mole fraction. Consequently, the graph displays a curve that arches upwards, lying consistently above the straight line that would connect the origin to the pure component vapor pressure.

The molar mass of solute $X$ is determined using the boiling point elevation equation $\Delta T_b = K_b \frac{w_2}{M_2 W_1}$. Substituting the given values $0.5 = 5.0 \frac{1.5}{M_2 \cdot 0.150}$ yields $M_2 = 100$ g mol$^{-1}$.

According to Raoult's law for a dilute solution, the relative lowering of vapour pressure is equal to the mole fraction of the solute, $\frac{\Delta p}{p^\circ} = \frac{n_2}{n_1}$. Calculating the moles gives $n_2 = \frac{1.5}{100} = 0.015$ mol and $n_1 = \frac{150}{300} = 0.5$ mol. Consequently, the relative lowering of vapour pressure is $\frac{0.015}{0.5} = 0.03$, which is equal to $3 \times 10^{-2}$.

The perxenate ion, $\mathrm{XeO}^{4-}_{6}$, adopts an octahedral geometry where the central xenon atom is bonded to six oxygen atoms. Each of these connections corresponds to a $\sigma$ bond, and since all available electrons are utilized in these linkages, there are 0 lone pairs remaining on the central atom. Consequently, the sum of the $\sigma$ bond-pairs and the lone pairs equals 6.

The osmotic pressure of an aqueous solution is calculated using the formula $\Pi = iCRT$, where $i$ is the number of ions produced upon complete dissociation, $C$ is the molar concentration, $R$ is the gas constant, and $T$ is the temperature. For $\mathrm{K}_4[\mathrm{Fe}(\mathrm{CN})_6]$, dissociation yields 5 ions per formula unit, resulting in an osmotic pressure of $\Pi_1 = 5 \times 0.1 \times RT = 0.5 RT$, while $\mathrm{FeCl}_3$ yields 4 ions per formula unit, resulting in $\Pi_2 = 4 \times 0.1 \times RT = 0.4 RT$. Because $\Pi_2 < \Pi_1$, the solution with the lower osmotic pressure is characterized as hypotonic. A semi-permeable membrane allows only the flow of solvent molecules, preventing the mixing of solute ions required to form a blue precipitate, and reverse osmosis requires an applied external pressure strictly greater than the osmotic pressure difference, which makes the other provided statements invalid.

Henry's law states that the partial pressure of a gas $p$ in equilibrium with a solution is directly proportional to its mole fraction $x$ in the solvent, represented by the equation $p = K_H x$, where $K_H$ is the Henry's law constant at a given temperature. Applying the logarithm to both sides of this equation results in $\log(p) = \log(x) + \log(K_H)$, which conforms to the linear equation $y = mx + c$. This mathematical relationship indicates that a plot of $\log(p)$ on the $y$-axis versus $\log(x)$ on the $x$-axis produces a straight line with a slope of 1 and a $y$-intercept equal to $\log(K_H)$.

Mole fraction depends on the ratio of moles of a specific component to the total moles in a mixture, requiring the conversion of mass to moles using molar mass. When a solution is defined by weight percentage, assuming a sample size of 100 g simplifies the arithmetic to working directly with grams. For a mixture of methanol and carbon tetrachloride, the molar masses are 32 g/mol and 154 g/mol respectively. With 30 g of methanol and 70 g of carbon tetrachloride in a 100 g solution, the moles of carbon tetrachloride are calculated as 70/154 and the moles of methanol as 30/32. Dividing the moles of carbon tetrachloride by the sum of moles of both components, denoted as $\frac{70/154}{(70/154) + (30/32)}$, results in approximately 0.33, confirming the first statement.

Positive deviation from Raoult's law arises when intermolecular forces between the solute and solvent are weaker than the cohesive forces within the pure components. In the mixture of methanol and carbon tetrachloride, the non-polar carbon tetrachloride molecules disrupt the strong hydrogen bonding network present in pure methanol. This reduced interaction between the unlike molecules allows them to escape more readily into the vapor phase, resulting in a higher total vapor pressure than predicted by Raoult's law for an ideal solution, which validates the second statement.

Osmotic pressure is a colligative property governed by the equation $\pi = \frac{wRT}{MV}$, where $w$ is the mass of the solute, $M$ is the molar mass, $V$ is the volume of the solution, $R$ is the gas constant, and $T$ is the temperature. For solution A, substituting $w = 1$ g, $M = 50000$ g mol$^{-1}$, $V = 0.5$ L, $R = 0.083$ L bar K$^{-1}$ mol$^{-1}$, and $T = 300$ K gives $x = \frac{1 \times 0.083 \times 300}{50000 \times 0.5} = 9.96 \times 10^{-4}$ bar. Similarly, solution B contains $w = 2$ g in $V = 1$ L, which yields $y = \frac{2 \times 0.083 \times 300}{50000 \times 1} = 9.96 \times 10^{-4}$ bar. When both solutions are combined, the resulting mixture has a total mass of 3 g and a total volume of 1.5 L, leading to the osmotic pressure $z = \frac{3 \times 0.083 \times 300}{50000 \times 1.5} = 9.96 \times 10^{-4}$ bar.

Raoult's Law for an ideal binary solution states that the total vapor pressure is given by $P_{total} = P_A^0 X_A + P_B^0 X_B$, where $P_A^0$ and $P_B^0$ are the vapor pressures of the pure components and $X_A$ and $X_B$ are their respective mole fractions. For the first mixture with 2 moles of liquid A and 3 moles of liquid B, the mole fractions are $X_A = 0.4$ and $X_B = 0.6$, leading to the equation $0.4 P_A^0 + 0.6 P_B^0 = 320$, which simplifies to $2 P_A^0 + 3 P_B^0 = 1600$. When 1 mole each of A and B is added, the new mixture contains 3 moles of A and 4 moles of B, resulting in mole fractions $X_A = 3/7$ and $X_B = 4/7$. This produces the second equation $\frac{3}{7} P_A^0 + \frac{4}{7} P_B^0 = 328.6$, which simplifies to $3 P_A^0 + 4 P_B^0 = 2300.2$. Solving the simultaneous linear equations $2 P_A^0 + 3 P_B^0 = 1600$ and $3 P_A^0 + 4 P_B^0 = 2300$ by multiplying the first by 3 and the second by 2 gives $6 P_A^0 + 9 P_B^0 = 4800$ and $6 P_A^0 + 8 P_B^0 = 4600$. Subtracting the second equation from the first yields $P_B^0 = 200$ mm Hg, and substituting this value back into the first equation results in $2 P_A^0 + 600 = 1600$, confirming $P_A^0 = 500$ mm Hg.

The osmotic pressure for an electrolyte solution is defined by the formula $\pi = iCRT$, where $i$ is the van't Hoff factor, $C$ is the molarity in $\mathrm{mol}\,\mathrm{L}^{-1}$, $R$ is the gas constant, and $T$ is the temperature. Given that NaCl dissociates completely into $\mathrm{Na}^+$ and $\mathrm{Cl}^-$, the van't Hoff factor is $i = 2$, and the molar mass of NaCl is $23 + 35.5 = 58.5\,\mathrm{g}\,\mathrm{mol}^{-1}$. Expressing molarity as the ratio of strength $S$ in $\mathrm{g}\,\mathrm{L}^{-1}$ to molar mass $M$, the equation becomes $12 = 2 \times \left(\frac{S}{58.5}\right) \times 0.08 \times 300$. This simplifies to $12 = \frac{48S}{58.5}$, which yields $S = \frac{12 \times 58.5}{48} = 14.625\,\mathrm{g}\,\mathrm{L}^{-1}$. Rounding to the nearest integer gives 15.

The elevation in boiling point is given by the formula $\Delta T_b = \frac{1000 \cdot K_b \cdot w_2}{M_2 \cdot w_1}$, where $w_2$ is the mass of the solute, $w_1$ is the mass of the solvent, and $K_b$ is the ebullioscopic constant.

Substituting the values for compound $PQ$ into this expression, $1.176 = \frac{1000 \cdot 5 \cdot 1}{M_{PQ} \cdot 50}$ leads to $M_{PQ} = \frac{100}{1.176} \approx 85 \text{ g mol}^{-1}$. Similarly, for compound $PQ_2$, $0.689 = \frac{1000 \cdot 5 \cdot 1}{M_{PQ_2} \cdot 50}$ yields $M_{PQ_2} = \frac{100}{0.689} \approx 145 \text{ g mol}^{-1}$.

Defining $M_P$ and $M_Q$ as the molar masses of elements $P$ and $Q$, the system of equations $M_P + M_Q = 85$ and $M_P + 2M_Q = 145$ is obtained. Subtracting the first from the second results in $M_Q = 60 \text{ g mol}^{-1}$, which gives $M_P = 85 - 60 = 25 \text{ g mol}^{-1}$.

Fluoro acetic acid acts as a weak electrolyte in aqueous solution, so the measured freezing point depression accounts for the partial ionization of the acid. Finding the dissociation constant involves determining the degree of ionization by evaluating the van't Hoff factor, which relates the observed freezing point change to the theoretical concentration.

The quantity of fluoro acetic acid in moles is calculated as $19.5/78 = 0.25$. Dissolving this in 0.5 kg of water creates a molality of $0.25/0.5 = 0.5$ mol kg$^{-1}$, which is treated as the molarity $C$ per the problem assumptions. Substituting these values into the colligative equation $\Delta T_f = i K_f m$ results in $1 = i \times 1.86 \times 0.5$, which simplifies to $i = 2/1.86 \approx 1.075$. Because the acid dissociates into two ions, the relation $i = 1 + \alpha$ provides a degree of ionization of $\alpha = 0.075$. Using the standard expression $K_a = \frac{C \alpha^2}{1-\alpha}$ with these values leads to $K_a = 0.5 \times \frac{(0.075)^2}{1-0.075}$, which equals $3 \times 10^{-3}$.

Henry's law is defined by the relationship $P = K_H \cdot \chi$, where $P$ is the partial pressure of the gas above the solution, $\chi$ is the mole fraction of the gas, and $K_H$ is the Henry's law constant. For an ideally dilute solution, the interactions between solute and solvent remain consistent, rendering $K_H$ a characteristic constant that does not change with concentration variations. Because $K_H$ is inherently dependent on the intermolecular forces between the specific solute and solvent molecules, it varies significantly when the solvent is changed, making the assertion that it remains the same across different solvents incorrect.

Osmotic pressure is determined by the total molar concentration of all solute particles in the solution, calculated using the formula $\pi = \frac{n_{total}RT}{V}$. The number of moles for each solute is $n_A = \frac{0.3\,\mathrm{g}}{60\,\mathrm{g}\,\mathrm{mol}^{-1}} = 0.005\,\mathrm{mol}$ and $n_B = \frac{0.9\,\mathrm{g}}{180\,\mathrm{g}\,\mathrm{mol}^{-1}} = 0.005\,\mathrm{mol}$, providing a total of $n_{total} = 0.01\,\mathrm{mol}$. Given a volume $V = 100\,\mathrm{mL} = 0.1\,\mathrm{L}$ and temperature $T = 300\,\mathrm{K}$, the final calculation is $\pi = \frac{0.01\,\mathrm{mol} \times 0.082\,\mathrm{L}\,\mathrm{atm}\,\mathrm{K}^{-1}\,\mathrm{mol}^{-1} \times 300\,\mathrm{K}}{0.1\,\mathrm{L}} = 2.46\,\mathrm{atm}$.

Henry's Law states that the partial pressure of a gas above a liquid is directly proportional to its mole fraction in the solution, represented by the equation $P_{\text{gas}} = K_H x_{\text{gas}}$.

First, calculate the partial pressure of nitrogen by multiplying its mole percentage in air by the total pressure: $P_{\text{N}_2} = 0.80 \times 10\text{ atm} = 8\text{ atm}$. Convert this pressure to mmHg to maintain unit consistency with the Henry's Law constant, resulting in $8 \times 760\text{ mmHg} = 6080\text{ mmHg}$.

Solve for the mole fraction of nitrogen in water by dividing the partial pressure by the Henry's Law constant:
$x_{\text{N}_2} = \frac{6080\text{ mmHg}}{6.5 \times 10^7\text{ mmHg}} = 9.35 \times 10^{-5}$