Solutions: 10 Years Solved PYQs

Correct Option: (C) Molarity

Reason: Molarity ($M = \frac{n}{V}$) is defined as moles of solute per litre of solution. Because volume expands or contracts with changes in temperature ($V \propto T$), molarity changes with temperature. Molality, mole fraction, and mass percentage depend purely on masses, which remain independent of temperature.

Definition: Parts per million (ppm) is defined as the parts of a solute component per million ($10^6$) parts of the total solution by mass:

Application: It is preferred when a solute is present in extremely minute trace quantities, such as quantifying atmospheric pollutants ($\text{SO}_2, \text{NO}_x$), dissolved chlorine in drinking water, or heavy metal toxicities.

Step 1: Determine masses: $20\%\ (w/w)$ means $20\text{ g}$ of $\text{KI}$ is dissolved in $(100 - 20) = 80\text{ g}$ of water ($\text{solvent}$). Moles of $\text{KI}$ ($n_{\text{KI}}$) = $\frac{20}{166} = 0.1205\text{ mol}$.

Step 2: Calculate Molality ($m$):

Step 3: Calculate Molarity ($M$): Volume = $\frac{\text{Mass}}{\text{Density}} = \frac{100}{1.202} = 83.19\text{ mL} = 0.08319\text{ L}$.

Molar mass: $\text{C}_2\text{H}_6\text{O}_2 = 62\text{ g mol}^{-1}$, $\text{H}_2\text{O} = 18\text{ g mol}^{-1}$.
In $100\text{ g}$ solution: $w_{\text{glycol}} = 20\text{ g}$, $w_{\text{water}} = 80\text{ g}$.
$n_{\text{glycol}} = \frac{20}{62} = 0.322\text{ mol}$,    $n_{\text{water}} = \frac{80}{18} = 4.444\text{ mol}$.

Henry's Law: At constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid surface: $p = K_H \cdot x$.

(i) Temperature: Gas solubility decreases as temperature rises because dissolution is exothermic ($\Delta_{\text{sol}}H < 0$), shifting equilibrium backward (Le Chatelier's Principle).

(ii) Pressure: Gas solubility increases with increasing pressure because compression increases the collision rate of gas molecules with the liquid surface.

Answer: Gas B is more soluble.

Reason: From Henry's law, $x = \frac{p}{K_H}$. Solubility ($x$) is inversely proportional to $K_H$. Because gas $B$ has a significantly lower $K_H$ value ($1.67\text{ kbar} < 40.3\text{ kbar}$), it dissolves to a much greater extent.

Dissolution of oxygen in water is exothermic ($\Delta_{\text{sol}}H < 0$). As temperature decreases, Henry's constant ($K_H$) decreases, resulting in higher solubility of oxygen in cold water. Cold water therefore provides a significantly higher dissolved oxygen concentration, facilitating easier respiration for aquatic life.

Cause: High pressure deep underwater forces large amounts of atmospheric nitrogen to dissolve in a diver's blood. When the diver surfaces rapidly, the pressure drops quickly, releasing dissolved nitrogen as gas bubbles in the bloodstream that block blood capillaries and pinch nerves.

Symptoms: Extreme pain, cramping, and risk of paralysis known as the bends.

Prevention: Breathing tanks are filled with air diluted with helium ($11.7\%\text{ He}, 56.2\%\text{ N}_2, 32.1\%\text{ O}_2$), as helium has exceptionally low solubility in blood.

At high altitudes, atmospheric pressure is substantially lower than at ground level. According to Henry's Law, the partial pressure of oxygen drops, reducing the concentration of oxygen dissolved in the climber's blood and tissues. This deficiency leads to muscular fatigue and cognitive confusion, termed Anoxia.

In $1\text{ L}$ of water ($1000\text{ g}$): $n_{\text{water}} = \frac{1000}{18} = 55.55\text{ mol}$. Since $n_{\text{N}_2} \ll n_{\text{water}}$:

Raoult's Law: $p_i = p_i^\circ \cdot x_i$.    Henry's Law: $p_i = K_H \cdot x_i$.

Both equations state that the partial vapour pressure of a component is directly proportional to its mole fraction in solution. They differ only in the proportionality constant: $p_i^\circ$ in Raoult's law and $K_H$ in Henry's law. When $K_H = p_i^\circ$, Henry's law becomes identical to Raoult's law.

Deviation: Positive Deviation ($\Delta_{\text{mix}}H > 0$, $\Delta_{\text{mix}}V > 0$).

Explanation: Pure ethanol molecules are held together by strong intermolecular hydrogen bonds. When acetone is added, its molecules insert themselves between ethanol molecules, breaking existing hydrogen bonds. Because the new ethanol-acetone attractive forces ($A-B$) are weaker than the original ethanol-ethanol forces ($A-A$), molecules escape more readily into vapour, resulting in higher total vapour pressure.

Correct Option: (B) Benzene + Toluene ($\text{C}_6\text{H}_6 + \text{C}_6\text{H}_5\text{CH}_3$)

Reason: Both are non-polar hydrocarbons of similar molecular geometry and polarity; hence $A-B$ intermolecular forces match $A-A$ and $B-B$ forces ($\Delta H_{\text{mix}} = 0, \Delta V_{\text{mix}} = 0$).

(a) Glucose is a non-volatile solute. In solution, glucose molecules occupy part of the liquid surface area, reducing the fraction of surface occupied by volatile water molecules. This lowers the evaporation rate and decreases vapour pressure.

(b) Azeotropes are binary liquid mixtures that boil at a constant temperature with identical compositions in both liquid and vapour phases. They cannot be separated by fractional distillation because both components boil off simultaneously in the same proportion.

1. Minimum Boiling Azeotrope: $95.4\%$ Ethanol and $4.6\%$ Water (by volume). Exhibits large positive deviation from Raoult's law. It boils at $351.15\text{ K}$, lower than pure ethanol ($351.5\text{ K}$) and pure water ($373.15\text{ K}$).

2. Maximum Boiling Azeotrope: $68\%$ Nitric Acid ($\text{HNO}_3$) and $32\%$ Water (by mass). Exhibits large negative deviation. It boils at $393.5\text{ K}$, higher than pure water ($373.15\text{ K}$) and pure $\text{HNO}_3$ ($359\text{ K}$).

Phenol and aniline form intermolecular hydrogen bonds between the phenolic hydrogen atom and the lone pair on the aniline nitrogen atom ($\text{C}_6\text{H}_5-\text{O}-\text{H}\cdots:\text{NH}_2-\text{C}_6\text{H}_5$). Because these newly formed solute-solvent attractions ($A-B$) are stronger than the pure solvent-solvent ($A-A$) and solute-solute ($B-B$) attractions, molecules escape less readily into the vapour phase, reducing total vapour pressure.

$n_{\text{glucose}} = \frac{30}{180} = 0.1667\text{ mol}$,    $n_{\text{water}} = \frac{500}{18} = 27.778\text{ mol}$.

$n_{\text{CHCl}_3} = \frac{25.5}{119.5} = 0.213\text{ mol}$,    $n_{\text{CH}_2\text{Cl}_2} = \frac{40.0}{85.0} = 0.471\text{ mol}$. Total moles = $0.684\text{ mol}$.
$x_{\text{CH}_2\text{Cl}_2} = \frac{0.471}{0.684} = 0.688$,    $x_{\text{CHCl}_3} = 1 - 0.688 = 0.312$.

On high mountains, reduced atmospheric pressure lowers the normal boiling point of water ($< 100^\circ\text{C}$), slowing down cooking. A pressure cooker raises internal pressure above $1\text{ atm}$, elevating the boiling point of water ($> 120^\circ\text{C}$) and cooking food much faster.

Definition: $K_b$ is the elevation in boiling point produced when 1 mole of a non-volatile, non-electrolyte solute is dissolved in $1\text{ kg}$ ($1000\text{ g}$) of a solvent (i.e., a $1\text{ molal}$ solution).

SI Unit: $\mathbf{\text{K kg mol}}^{-1}$ (or $^\circ\text{C kg mol}^{-1}$).

Salt dissolves into the thin surface liquid film on ice, creating an electrolytic solution that depresses the freezing point of water well below $0^\circ\text{C}$ (down to $-15^\circ\text{C}$ for $\text{NaCl}$ and $-30^\circ\text{C}$ for $\text{CaCl}_2$). At typical winter temperatures, snow melts into liquid solution and drains away, clearing the road.

(i) Food Preservation: Concentrated brine and sugar syrups are hypertonic to bacterial and fungal cells. Water is drawn out of microbes via exosmosis (plasmolysis), dehydrating and killing them.

(ii) Edema: Excess sodium elevates osmotic pressure in extracellular tissue fluids, drawing water out of blood capillaries into surrounding tissue spaces via osmosis, causing physical swelling (edema).

$\Delta T_{f,\text{sugar}} = 273.15 - 271.00 = 2.15\text{ K}$.
$2.15 = K_f \times \frac{5 \times 1000}{342 \times 95} \implies K_f = \frac{2.15 \times 342 \times 95}{5000} = 13.97\text{ K kg mol}^{-1}$.

For $5\%$ Glucose solution:
$\Delta T_{f,\text{glucose}} = 13.97 \times \frac{5 \times 1000}{180 \times 95} = 4.08\text{ K}$.
$T_f = 273.15 - 4.08 = \mathbf{269.07\text{ K}}$.

When an external hydrostatic pressure greater than the osmotic pressure ($P > \Pi$) is applied to the saline solution side, pure water molecules are forced backward through a semipermeable membrane (such as cellulose acetate) into the pure water reservoir, leaving dissolved salts and impurities behind.

$\text{BaCl}_2 \rightarrow \text{Ba}^{2+} + 2\text{Cl}^-$. Since 1 formula unit produces $1 + 2 = 3$ ions in solution, $\mathbf{i = 3}$.

$\Delta T_{f,\text{obs}} = 0 - (-0.068) = 0.068\text{ K}$.
$\Delta T_{f,\text{calc}} = K_f \times m = 1.86 \times 0.01 = 0.0186\text{ K}$.

For $\text{AlCl}_3 \rightarrow \text{Al}^{3+} + 3\text{Cl}^-$ ($n = 4$):

Elevation in boiling point depends on effective particle count ($i \times C$):
Glucose ($i=1, iC=0.1$) < $\text{NaCl}$ ($i=2, iC=0.2$) < $\text{BaCl}_2$ ($i=3, iC=0.3$) < $\text{AlCl}_3$ ($i=4, iC=0.4$).

Order: $0.1\text{ M Glucose} < 0.1\text{ M NaCl} < 0.1\text{ M BaCl}_2 < 0.1\text{ M AlCl}_3$

$m = \frac{2.0 \times 1000}{122 \times 25.0} = 0.6557\text{ mol kg}^{-1}$.
$\Delta T_{f,\text{calc}} = 4.9 \times 0.6557 = 3.213\text{ K}$.
$i = \frac{1.62}{3.213} = 0.5042$.

For complete dissociation of $\text{CaCl}_2 \rightarrow \text{Ca}^{2+} + 2\text{Cl}^-$, $i = 3$.
$m = \frac{10.0 \times 1000}{111 \times 200} = 0.4505\text{ mol kg}^{-1}$.

$\text{MgBr}_2 \rightarrow \text{Mg}^{2+} + 2\text{Br}^- \implies i = 3$.
$m = \frac{10.50 \times 1000}{184 \times 200} = 0.2853\text{ mol kg}^{-1}$.

$m = \frac{19.5 \times 1000}{78 \times 500} = 0.50\text{ mol kg}^{-1}$.
$\Delta T_{f,\text{calc}} = K_f \times m = 1.86 \times 0.50 = 0.93\text{ K}$.

For $\text{CH}_2\text{FCOOH} \rightleftharpoons \text{CH}_2\text{FCOO}^- + \text{H}^+$ ($n = 2$):
$\alpha = \frac{i - 1}{n - 1} = \frac{1.0753 - 1}{2 - 1} = \mathbf{0.0753}$.

(i) $i = 1$: Normal behavior. Solute undergoes neither dissociation nor association in the solvent (e.g., Glucose, Urea, Sucrose in water).

(ii) $i > 1$: Dissociation. Solute breaks apart into two or more ions, increasing total particle concentration (e.g., $\text{NaCl}, \text{CaCl}_2, \text{K}_2\text{SO}_4$ in water).

(iii) $i < 1$: Association. Solute molecules assemble into dimers or multimers, decreasing total particle count (e.g., Benzoic acid or Ethanoic acid in benzene).