Solutions – JEE Main ChemistryPractice Questions & PYQs

According to Raoult's Law, the relative lowering of vapour pressure (RLVP) is defined as the mole fraction of the solute, $\frac{n_2}{n_1 + n_2}$, which approximates to $\frac{n_2}{n_1}$ when the number of moles of solute ($n_2$) is very small compared to the moles of solvent ($n_1$). Substituting $n_1 = \frac{W_1}{M_1}$ (where $W_1$ is the solvent mass in grams and $M_1$ is its molar mass) and the definition of molality $m' = \frac{n_2 \times 1000}{W_1}$ into the RLVP expression gives $\text{RLVP} = \frac{n_2 \cdot M_1}{W_1} = \frac{m' \cdot M_1}{1000}$.

Since the molality ($m' = 5$) is identical for both solutions, the RLVP is directly proportional to the molar mass of the solvent, resulting in the ratio $\frac{\text{RLVP}_X}{\text{RLVP}_Y} = \frac{M_X}{M_Y}$. Given $M_X = \frac{3}{4} M_Y$, substituting this into the ratio yields $m = \frac{3}{4}$.

The relative lowering of vapor pressure is expressed by the equation $\frac{P^\circ - P_s}{P_s} = \frac{n_B}{n_A}$, where $P^\circ$ is the pure vapor pressure of the solvent, $P_s$ is the vapor pressure of the solution, and $n_A$ and $n_B$ represent the moles of solvent and solute, respectively. Reducing the vapor pressure by 75% implies that the final vapor pressure is 25% of the original, so $P_s = 0.25 P^\circ$, resulting in a ratio of $\frac{P^\circ - 0.25 P^\circ}{0.25 P^\circ} = \frac{0.75 P^\circ}{0.25 P^\circ} = 3$. Given that the mass of octane ($C_8H_{18}$) is 114 g and its molar mass is 114 g mol$^{-1}$, the moles of solvent $n_A$ equals 1 mol. Substituting these values gives $3 = \frac{n_B}{1}$, so $n_B = 3 \text{ mol}$, and the mass of the solute required is $3 \text{ mol} \times 50 \text{ g mol}^{-1} = 150 \text{ g}$.

The freezing point depression of a solution depends on the total concentration of solute particles, which is adjusted for electrolytes by the Van't Hoff factor. Since $\mathrm{Na}_2\mathrm{SO}_4$ dissociates in water according to $\mathrm{Na}_2\mathrm{SO}_4 \rightarrow 2\mathrm{Na}^{+} + \mathrm{SO}_4^{2-}$, there are 3 particles produced per formula unit. Given a degree of ionization of 0.815, the Van't Hoff factor is calculated as $i = 1 + (3 - 1) \times 0.815 = 2.63$.

The molar mass of $\mathrm{Na}_2\mathrm{SO}_4$ is $2 \times 23 + 32 + 4 \times 16 = 142 \text{ g mol}^{-1}$. Applying the freezing point depression formula $\Delta T_f = i \times K_f \times m$, we substitute the known values into $3.82 = 2.63 \times 1.86 \times \frac{5 \times 1000}{142 \times x}$. Rearranging this equation to solve for $x$ gives $x = \frac{2.63 \times 1.86 \times 5000}{142 \times 3.82}$, which results in approximately 45 g.

When acetic acid dissolves in benzene, its molecules associate to form dimers, meaning the solute particles decrease in number, which affects the solution's colligative properties. The deviation from ideal behavior is accounted for by the van't Hoff factor, $i$, which for a dimerization process is defined as $i = 1 - \frac{\alpha}{2}$, where $\alpha$ represents the degree of association. The relationship between the observed freezing point depression and the solute concentration is given by the formula $\Delta T_f = i \cdot K_f \cdot m$.

To determine the degree of association, we first calculate the molality $m$ of the solution, which is the moles of solute divided by the mass of the solvent in kilograms. With $0.2 \text{ g}$ of acetic acid having a molar mass of $60 \text{ g/mol}$ dissolved in $0.020 \text{ kg}$ of benzene, the molality is $m = \frac{0.2/60}{0.020} = 0.1667 \text{ mol/kg}$. Inserting the provided freezing point depression $\Delta T_f = 0.45^\circ\text{C}$ and the cryoscopic constant $K_f = 5.12 \text{ K kg mol}^{-1}$ into the equation, we have $0.45 = (1 - \frac{\alpha}{2}) \cdot 5.12 \cdot 0.1667$. Solving this for the van't Hoff factor yields $i = \frac{0.45}{0.8533} \approx 0.527$. By rearranging $1 - \frac{\alpha}{2} = 0.527$, we find that $\frac{\alpha}{2} = 0.473$, leading to $\alpha \approx 0.946$. This indicates that the percentage association of acetic acid in benzene is $94.6\%$.

Raoult's Law and Dalton's Law of partial pressures govern the relationship between liquid composition and vapor composition: $P_i = x_i P^o_i$ and $y_i = \frac{P_i}{P_{total}}$. Calculating the molar masses gives $M(CH_2Cl_2) = 12 + 2(1) + 2(35.5) = 85 \text{ g/mol}$ and $M(CHCl_3) = 12 + 1 + 3(35.5) = 119.5 \text{ g/mol}$. The number of moles for each component is $n(CH_2Cl_2) = \frac{8.5}{85} = 0.1 \text{ mol}$ and $n(CHCl_3) = \frac{11.95}{119.5} = 0.1 \text{ mol}$, resulting in liquid mole fractions of $x(CH_2Cl_2) = 0.5$ and $x(CHCl_3) = 0.5$.

Partial pressures are determined as $P(CH_2Cl_2) = 0.5 \times 415 = 207.5 \text{ mmHg}$ and $P(CHCl_3) = 0.5 \times 200 = 100 \text{ mmHg}$. Summing these provides the total vapor pressure of the solution, $P_{total} = 207.5 + 100 = 307.5 \text{ mmHg}$. Consequently, the mole fraction of $CHCl_3$ in the vapor phase is $y(CHCl_3) = \frac{100}{307.5} = 0.325$.

Henry's Law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the solution, meaning an increase in pressure forces more gas to dissolve. This principle allows us to relate the initial solubility and pressure to the final state using the proportionality constant $S_1 / P_1 = S_2 / P_2$.

Given an initial solubility of 0.01 g/L at 500 torr and a target pressure of 750 torr, we can set up the following equation to find the new solubility $S_2$:
$\frac{0.01}{500} = \frac{S_2}{750}$
Solving for the unknown variable $S_2$ requires multiplying 0.01 by 750 and then dividing by 500, which results in 0.015 g/L.

The van't Hoff factor measures the extent of solute dissociation in a solution, defined by the formula
$i = 1 + (n - 1)\alpha$
where $n$ is the total number of ions produced per formula unit and $\alpha$ is the degree of dissociation. For the salt $MX_2$, each formula unit dissociates into one $M^{2+}$ ion and two $X^-$ ions, meaning the total number of ions produced is $n = 3$.

Substituting the known values into the equation yields $2 = 1 + (3 - 1)\alpha$. This expression simplifies to $2 = 1 + 2\alpha$, which rearranges to $2\alpha = 1$, resulting in a degree of dissociation of $\alpha = 0.50$.

The reduction of vapor pressure when a non-volatile solute is added to a solvent is defined by Raoult's Law, which states that the relative lowering of vapor pressure is equal to the mole fraction of the solute. This relationship is expressed as:

$\frac{p^0 - p}{p^0} = \frac{n_{glucose}}{n_{glucose} + n_{water}}$

Calculating the moles of each substance reveals the proportions in the solution. With a molar mass of $180 \text{ g/mol}$, the $18 \text{ g}$ of glucose corresponds to $0.1 \text{ mol}$. Water, having a molar mass of $18 \text{ g/mol}$, provides $9.9 \text{ mol}$ from the $178.2 \text{ g}$ sample. The mole fraction of glucose is therefore $\frac{0.1}{0.1 + 9.9}$, which simplifies to $0.01$. Using the standard vapor pressure of water of $760 \text{ torr}$, the pressure lowering is $0.01 \times 760 = 7.6 \text{ torr}$. The final vapor pressure of the solution is found by subtracting this decrease from the pure solvent's vapor pressure, resulting in $760 - 7.6 = 752.4 \text{ torr}$.

Raoult's Law describes how the vapor pressure of a solution decreases when a non-volatile solute is added, stating that the relative lowering of vapor pressure is equal to the mole fraction of the solute, expressed as $\frac{P^0 - P}{P} = \frac{n_{\text{solute}}}{n_{\text{solvent}}}$. To apply this to the given problem, the relative lowering is $\frac{185 - 183}{183}$, using the pure vapor pressure of 185 torr and the solution vapor pressure of 183 torr. The solvent, acetone, has a mass of 100 g and a molar mass of 58 g mol⁻¹, meaning the number of moles of solvent is $\frac{100}{58}$. With 1.2 g of solute, the number of moles is $\frac{1.2}{M}$, leading to the equation $\frac{2}{183} = \frac{1.2/M}{100/58}$. Solving for the molar mass $M$ results in $M = \frac{1.2 \times 58 \times 183}{2 \times 100}$, which gives a value of approximately 64 g mol⁻¹.

Osmotic pressure is a colligative property defined by the relationship $\pi = iMRT$, where $i$ represents the van 't Hoff factor, $M$ is the molarity, and $R$ and $T$ are the gas constant and temperature, respectively. Since the temperature remains constant for all samples, the relative osmotic pressure of each solution depends entirely on the product $iM$.

For the non-electrolyte $\mathrm{C_2H_5OH}$, which does not dissociate, the van 't Hoff factor is 1, resulting in a product of $1 \times 0.500 = 0.500$. The strong electrolyte $\mathrm{KBr}$ dissociates into two ions, yielding $i=2$, so its product is $2 \times 0.250 = 0.500$. Similarly, the salt $\mathrm{Mg_3(PO_4)_2}$ dissociates into five ions-three magnesium ions and two phosphate ions-which provides $i=5$ and results in a calculation of $5 \times 0.100 = 0.500$. Finally, $\mathrm{Na_3PO_4}$ dissociates into four ions, giving $i=4$, which leads to $4 \times 0.125 = 0.500$. Because the value of $iM$ is identical for every solution, they all exert the same osmotic pressure.

Freezing point depression illustrates how the temperature at which a solvent solidifies drops as non-volatile solute particles are introduced. This behavior is quantified by the equation $\Delta T_f = i \times K_f \times m$, where $\Delta T_f$ is the shift in freezing temperature, $i$ is the van't Hoff factor, $K_f$ is the cryoscopic constant, and $m$ is the molality of the solution.

Because ethylene glycol is a non-electrolyte, its van't Hoff factor is 1. To depress the freezing point of 1.0 kg of water by 2.8 degrees Celsius using this substance, which has a molar mass of 62 g/mol, we set up the equation $2.8 = 1 \times 1.86 \times \frac{x}{62}$, where $x$ represents the mass of the solute in grams. Isolating $x$ requires multiplying 2.8 by 62 and dividing the product by 1.86. Performing this calculation yields 93 g as the mass of ethylene glycol required for the radiator.

The freezing point depression is a colligative property defined by the relationship $\Delta T_f = K_f \times m$, where $\Delta T_f$ is the decrease in freezing point, $K_f$ is the cryoscopic constant, and $m$ is the molality of the solution. To prevent 4 kg of water from freezing at -6 degrees Celsius, the solution requires a freezing point depression of 6 degrees Celsius compared to the normal freezing point of pure water.

Molality is defined as the number of moles of solute per kilogram of solvent. Since the moles of ethylene glycol are calculated as the mass of the solute in grams, denoted by w, divided by its molar mass of 62 g mol^-1, the expression for the freezing point depression can be written as follows:

$\Delta T_f = \frac{K_f \times w}{\text{Molar mass} \times \text{Mass of solvent}}$

Substituting the known values into this equation gives:

$6 = \frac{1.86 \times w}{62 \times 4}$

Solving this equation for the mass w involves rearranging the terms as $w = \frac{6 \times 62 \times 4}{1.86}$, which results in 800 grams of ethylene glycol. This mass is the amount required to effectively maintain the target freezing point for the given quantity of water.

The van't Hoff factor represents the ratio of total particles at equilibrium to the initial formula units, with the degree of dissociation $\alpha$ tracking the extent of ion formation. For the reaction $A_x B_y \rightleftharpoons x A^{y+} + y B^{x-}$, setting an initial amount of one mole results in $1 - \alpha$ moles of electrolyte remaining and $x \alpha + y \alpha$ moles of product ions at equilibrium. Summing these components to determine the van't Hoff factor gives $i = 1 + \alpha(x + y - 1)$, which simplifies upon rearranging to the final relationship:

$\alpha = \frac{i-1}{x+y-1}$

Colligative properties like freezing point depression depend on the total number of solute particles in a solution. Because sodium sulphate is an electrolyte that dissociates completely in water, we must use the van't Hoff factor to account for the ions effectively present.

The dissociation reaction $\mathrm{Na}_2\mathrm{SO}_4 \rightarrow 2\mathrm{Na}^{+} + \mathrm{SO}_4^{2-}$ shows that each unit splits into three ions, giving a van't Hoff factor of $i = 3$. With a molality of 0.01 mol kg$^{-1}$ and a freezing point depression constant of 1.86 K kg mol$^{-1}$, we apply the equation $\Delta T_f = i \cdot K_f \cdot m$. Plugging in these values gives $\Delta T_f = 3 \times 1.86 \times 0.01$, resulting in 0.0558 K.

Raoult's Law dictates that the total vapor pressure of an ideal mixture is the sum of the partial pressures exerted by each individual component, where each partial pressure is the product of the component's pure vapor pressure and its mole fraction in the mixture. This relationship is defined by $P_{\text{total}} = P^\circ_{\text{heptane}} x_{\text{heptane}} + P^\circ_{\text{octane}} x_{\text{octane}}$.

Calculating the molar amounts for each liquid reveals that heptane provides $n_{\text{heptane}} = \frac{25.0}{100} = 0.25 \text{ mol}$, while octane provides $n_{\text{octane}} = \frac{35}{114} \approx 0.3 \text{ mol}$. Summing these provides a total of $0.55 \text{ mol}$ in the mixture. The resulting mole fractions are $\frac{0.25}{0.55}$ for heptane and $\frac{0.3}{0.55}$ for octane. Substituting these values into the pressure equation leads to $105 \text{kPa} \times \left( \frac{0.25}{0.55} \right) + 45 \text{kPa} \times \left( \frac{0.3}{0.55} \right)$. Performing the arithmetic yields $\frac{26.25 + 13.5}{0.55}$, which results in a final pressure of $72.0 \text{kPa}$.

Raoult's Law states that the total vapor pressure of an ideal solution is determined by the weighted sum of the vapor pressures of its pure components, where the weights are their respective mole fractions in the liquid phase. In the initial mixture containing 1 mole of X and 3 moles of Y, the mole fraction of X is $1/4$ and the mole fraction of Y is $3/4$, leading to the relation $550 = P_X^\circ \left(\frac{1}{4}\right) + P_Y^\circ \left(\frac{3}{4}\right)$, which simplifies to $P_X^\circ + 3P_Y^\circ = 2200$. When an additional mole of Y is added, the total number of moles becomes five, resulting in mole fractions of $1/5$ for X and $4/5$ for Y, and a total vapor pressure of $560\,\text{mm}\,\text{Hg}$. This second condition is expressed as $560 = P_X^\circ \left(\frac{1}{5}\right) + P_Y^\circ \left(\frac{4}{5}\right)$, or $P_X^\circ + 4P_Y^\circ = 2800$. Subtracting the first equation from the second directly gives the vapor pressure of pure Y as $P_Y^\circ = 600\,\text{mm}\,\text{Hg}$. Substituting this result back into the first equation yields $P_X^\circ + 3(600) = 2200$, confirming the vapor pressure of pure X is $400\,\text{mm}\,\text{Hg}$.

Whether a binary mixture behaves ideally depends on whether the intermolecular forces between the mixed components match the strength of the forces within the pure substances. Mixing n-heptane with ethanol disrupts the original cohesive forces because the new interactions between the different molecules are weaker than the n-heptane-n-heptane and ethanol-ethanol interactions found in their pure states. Since these reduced attractive forces allow more molecules to escape into the vapor phase, the total vapor pressure increases beyond what Raoult's law predicts, confirming that the solution is non-ideal and exhibits a positive deviation.

A liquid solution boils when its total vapour pressure equates to the external atmospheric pressure. In this case, the boiling point is 80 °C at 1 atm, which is equivalent to 760 mm Hg, allowing us to determine the mole fraction of component A by utilizing Raoult's law. The total pressure of the mixture is the sum of the partial pressures contributed by each component, expressed by the formula $P_T = P_A^{\circ} X_A + P_B^{\circ} X_B$

Substituting the given pure vapour pressures of 520 mm Hg and 1000 mm Hg, while acknowledging that the sum of the mole fractions $X_A + X_B$ must equal 1, leads to the equation $760 = 520 X_A + 1000(1 - X_A)$ which simplifies to $760 = 520 X_A + 1000 - 1000 X_A$. Rearranging the variables to combine the terms results \in $480 X_A = 240$ and solving for $X_A$ gives 0.5, which corresponds to 50 mol percent.

Adding a non-volatile solute like glucose reduces the vapor pressure of a solvent, a principle governed by Raoult's law. This relationship dictates that the ratio of the change in vapor pressure to the vapor pressure of the solution is equal to the ratio of the number of moles of solute to the number of moles of solvent.

For this system, the amount of glucose is 0.1 mol, calculated from 18 g divided by its molar mass of 180 g/mol, and the amount of water is 9.9 mol, from 178.2 g divided by 18 g/mol. Using the expression $\frac{P^{\circ} - P_s}{P_s} = \frac{n_{\text{glucose}}}{n_{\text{water}}}$ with an initial vapor pressure $P^{\circ}$ of 17.5 mm Hg results \in the equation $\frac{17.5 - P_s}{P_s} = \frac{0.1}{9.9}$. Rearranging to solve for $P_s$ leads to $17.5 - P_s = \frac{0.1}{9.9} P_s$, which ultimately yields a final solution vapor pressure of 17.325 mm Hg.

Isotonic solutions share identical osmotic pressure at the same temperature. Given the relationship $\pi = CRT$, this condition implies that the molar concentrations of the two solutions must be equal. When volumes are consistent, this equality allows us to compare the molar ratios of solute directly for both the unknown substance and the urea.

The urea solution concentration is determined by dividing its mass of 1.5 g by its molar mass of 60 g/mol. Similarly, the concentration of the unknown substance is calculated using its mass of 5.25 g and unknown molar mass $M$. Equating the molar concentrations yields the following equation:
$\frac{5.25}{M} = \frac{1.5}{60}$
Solving this expression for $M$ involves multiplying 5.25 by 60 and dividing by 1.5, which results in a molar mass of 210.0 g/mol.