JEE Solutions Questions

The conductometric titration curve represents the titration of a mixture of a strong acid (HCl) and a weak acid ($CH_3COOH$) with a strong base (NaOH).

The conductance of the solution depends on the number of ions present and their ionic mobility. Since the mobility of $$H^+$$ ions is exceptionally high, the conductance changes characteristically during the titration.

Initially, in segment AB, HCl is neutralized first because it is completely dissociated in solution. The highly mobile $$H^+$$ ions are replaced by the comparatively less mobile $$Na^+$$ ions from NaOH, causing a sharp decrease in conductance.

Point B, corresponding to $$2.0\ \text{mL}$$ of NaOH added, represents the equivalence point of HCl.

Beyond point B, in segment BC, NaOH starts neutralizing acetic acid:

$$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O.$$

Since acetic acid is only weakly ionized whereas sodium acetate is completely dissociated, the number of ions in solution increases, producing a gradual increase in conductance.

Point C, corresponding to $$5.0\ \text{mL}$$ of NaOH added, represents the complete neutralization of acetic acid.

After point C, any additional NaOH contributes free $$OH^-$$ ions to the solution. Because $$OH^-$$ ions possess high mobility, the conductance rises sharply in segment CD.

Hence:

• Option (C) is incorrect because HCl is neutralized before $$CH_3COOH$$.
• Option (D) is incorrect because point C corresponds to the equivalence point of $$CH_3COOH$$, while HCl is neutralized at point B.

Now calculate the concentrations.

Given,

$$V_{\text{mixture}}=40\ \text{mL},$$

$$M_{\text{NaOH}}=0.1\ \text{M}.$$

For HCl,

$$M_{\text{HCl}}\times40=0.1\times2.0,$$

so

$$M_{\text{HCl}}=\frac{0.2}{40}=0.005\ \text{M}.$$

For acetic acid, the volume of NaOH consumed is

$$5.0-2.0=3.0\ \text{mL}.$$

Therefore,

$$M_{CH_3COOH}\times40=0.1\times3.0,$$

giving

$$M_{CH_3COOH}=\frac{0.3}{40}=0.0075\ \text{M}.$$

Thus,

• $$[HCl]=0.005\ \text{M}$$
• $$[CH_3COOH]=0.0075\ \text{M}$$

Hence, the correct statement is that the concentration of HCl is

$$0.005\ \text{M},$$

so the correct answer is option (B).

To identify the correct conditions for reverse osmosis in this setup, we must understand the principles of osmosis, applied pressure, and membrane permeability.

The concentration of the solution in Chamber 1 is greater than that in Chamber 2, i.e.,

$$c_1 > c_2.$$

Under natural osmotic conditions, the solvent flows from the region of lower solute concentration (Chamber 2) to the region of higher solute concentration (Chamber 1) through a semipermeable membrane. The osmotic pressure $$\pi$$ is the minimum pressure that must be applied to the concentrated solution to stop this natural flow of solvent.

In reverse osmosis, the solvent is forced to move in the opposite direction, from the concentrated solution to the dilute solution. To achieve this, pressure must be applied on the concentrated side (Chamber 1), and the applied pressure must be greater than the osmotic pressure. Hence, the required condition is

$$p_1 > \pi.$$

The membrane separating the two chambers must also be a semipermeable membrane (SPM), which allows only solvent molecules to pass through while preventing the passage of solute particles. Cellophane and parchment paper act as semipermeable membranes, whereas an ordinary porous partition allows both solute and solvent to pass through and therefore cannot produce reverse osmosis.

Evaluating the given statements:

• (A) Cellophane with $$p_1 > \pi$$: Correct. Cellophane acts as a semipermeable membrane, and applying pressure greater than the osmotic pressure on the concentrated side produces reverse osmosis.
• (B) Porous partition with $$p_2 > \pi$$: Incorrect. A porous partition is not semipermeable, and applying pressure on the dilute side cannot produce reverse osmosis.
• (C) Parchment paper with $$p_1 > \pi$$: Correct. Parchment paper behaves as a semipermeable membrane, and the applied pressure satisfies the condition required for reverse osmosis.
• (D) Cellophane with $$p_2 > \pi$$: Incorrect. Although cellophane is a suitable membrane, applying pressure on the dilute side does not reverse the direction of solvent flow.

Therefore, the correct statements are A and C only.

Hence, the correct answer is

$$\boxed{\text{A and C only}}.$$

The relation between moles, molarity and volume is $$n = M \times V$$, where $$n$$ is in moles, $$M$$ in mol L$$^{-1}$$ and $$V$$ in litres.

Step 1 • Moles present before mixing
NaOH: $$M = 2\,\text{M},\; V = 10\,\text{mL}=0.01\,\text{L}$$
$$n_{NaOH} = 2 \times 0.01 = 0.02\,\text{mol}$$

HCl: $$M = 1\,\text{M},\; V = 20\,\text{mL}=0.02\,\text{L}$$
$$n_{HCl} = 1 \times 0.02 = 0.02\,\text{mol}$$

Step 2 • Neutralisation reaction
$$NaOH + HCl \rightarrow NaCl + H_2O$$
Since $$n_{NaOH}=n_{HCl}=0.02\,\text{mol}$$, both are completely consumed and we obtain $$0.02\,\text{mol}$$ NaCl.

Step 3 • Concentration of NaCl in the neutral mixture
Total volume after reaction $$=10\,\text{mL}+20\,\text{mL}=30\,\text{mL}=0.03\,\text{L}$$.
$$M_{NaCl} = \frac{0.02}{0.03} \approx 0.667\,\text{M}$$

Step 4 • Portion taken to the volumetric flask
Volume taken $$=10\,\text{mL}=0.01\,\text{L}$$.
Moles of NaCl transferred $$=0.667 \times 0.01 \approx 6.67\times10^{-3}\,\text{mol}$$.

Step 5 • HCl already present in the flask
The flask initially contains $$2\,\text{mol}$$ of HCl.

Step 6 • Final dilution to 100 mL
Final volume $$=100\,\text{mL}=0.1\,\text{L}$$.

Molarity of HCl:
$$M_{HCl} = \frac{2}{0.1} = 20\,\text{M}$$

Molarity of NaCl:
$$M_{NaCl} = \frac{6.67\times10^{-3}}{0.1} \approx 0.067\,\text{M}$$

Step 7 • Nature of the solution
The solution is a strongly acidic solution of $$20\,\text{M}$$ HCl (with a small amount of NaCl that does not affect acidity).

Hence the correct description is Option B - 20 M HCl solution.

For an ionic solid, the heat (enthalpy) of solution, $$\Delta H_{\text{sol}}$$, is obtained in two steps:
  1. Lattice dissociation: the crystal breaks into gaseous ions. This requires the lattice dissociation energy, $$\Delta H_{\text{lattice(diss)}}$$, which is numerically equal and opposite to the lattice energy of formation.
  2. Hydration: the gaseous ions get surrounded by water molecules, releasing their hydration energies.

Given data (sign conventions):
  • Hydration energy of $$K^+ = -x$$ kJ mol$$^{-1}$$ (exothermic)
  • Hydration energy of $$Cl^- = -y$$ kJ mol$$^{-1}$$ (exothermic)
  • Lattice energy of formation of KCl $$= -z$$ kJ mol$$^{-1}$$ (exothermic for formation from gaseous ions)

Step 1: Lattice dissociation energy
Lattice energy of formation is $$-z$$, so the energy required to dissociate the lattice is the opposite sign:
$$\Delta H_{\text{lattice(diss)}} = +z \text{ kJ mol}^{-1}$$

Step 2: Hydration of the ions
$$\Delta H_{\text{hydration}} = (-x) + (-y) = -(x + y) \text{ kJ mol}^{-1}$$

Overall heat of solution
$$\Delta H_{\text{sol}} = \Delta H_{\text{lattice(diss)}} + \Delta H_{\text{hydration}}$$
$$\Delta H_{\text{sol}} = \bigl(+z\bigr) + \bigl[-(x + y)\bigr]$$
$$\Delta H_{\text{sol}} = z - (x + y)$$

Therefore the heat of solution of KCl is $$z - (x + y)$$ kJ mol$$^{-1}$$.

Hence the correct option is Option C.

For a gas dissolved in water, Henry’s law is written as $$p = K_H \, x$$, where
$$p$$ = partial pressure of the gas above the solution,
$$x$$ = mole-fraction of the gas in the liquid phase,
$$K_H$$ = Henry’s law constant.

The solubility of the gas is inversely proportional to $$K_H$$: a lower $$K_H$$ means a larger mole-fraction (greater solubility), while a higher $$K_H$$ means a smaller mole-fraction (lower solubility).

Dissolution of almost every common gas in water is an exothermic process (heat is evolved). For an exothermic equilibrium the Le-Châtelier principle predicts that increasing temperature shifts the equilibrium in the direction that absorbs heat, i.e. towards the gaseous state. Hence the solubility of the gas decreases as temperature rises.

Mathematically, the temperature dependence of $$K_H$$ is obtained from the van’t Hoff relation:

$$\frac{d(\ln K_H)}{dT} = \frac{\Delta H}{R\,T^{2}} \qquad -(1)$$

Here $$\Delta H$$ is the enthalpy change for dissolution. Because $$\Delta H$$ is negative (exothermic), the right-hand side of $$(1)$$ is negative. Integrating, $$\ln K_H$$ increases with temperature, so $$K_H$$ itself increases monotonically with temperature.

Therefore a correct plot of $$K_H$$ versus temperature must show an increasing (upward-rising) curve.

Among the given alternatives, only Option D depicts $$K_H$$ steadily increasing as the temperature rises. All other graphs show either a decrease or a non-monotonic behaviour, contradicting the thermodynamic prediction.

Hence, the correct choice is:
Option D which is: $$K_H$$ increases with temperature.

Statement I: NaCl is added to ice at 0°C in the ice cream box to prevent melting of ice cream.

When NaCl is added to ice, the freezing point is depressed below 0°C. This means the ice-salt mixture can reach temperatures below 0°C, which helps keep the ice cream frozen. So NaCl does help prevent melting of ice cream. Statement I is TRUE.

Statement II: On addition of NaCl to ice at 0°C, there is a depression in freezing point.

This is a standard colligative property. Adding a solute (NaCl) depresses the freezing point of water. Statement II is TRUE.

Both statements are true. Statement II provides the reason for Statement I.

The correct answer is Option 4: Both Statement I and Statement II are true.

According to Raoult's law, the partial vapour pressure of each volatile component in an ideal solution is equal to the vapour pressure of the pure component multiplied by its mole fraction in the solution.

Given:

• Moles of liquid $$A = 1$$ mole
• Moles of liquid $$B = 3$$ moles
• Vapour pressure of pure $$A = 200\ \text{mm Hg}$$
• Total vapour pressure of the solution $$= 500\ \text{mm Hg}$$

The total number of moles in the solution is

$$n_{\text{total}} = n_A + n_B = 1 + 3 = 4$$

The mole fractions are therefore

$$x_A = \frac{1}{4} = 0.25$$

$$x_B = \frac{3}{4} = 0.75$$

Using Raoult's law,

$$P_{\text{total}} = P_A^\circ x_A + P_B^\circ x_B$$

Substituting the given values,

$$500 = (200 \times 0.25) + (P_B^\circ \times 0.75)$$

$$500 = 50 + 0.75P_B^\circ$$

$$450 = 0.75P_B^\circ$$

$$P_B^\circ = \frac{450}{0.75} = 600\ \text{mm Hg}$$

To determine the least volatile component, compare the vapour pressures of the pure liquids.

• $$P_A^\circ = 200\ \text{mm Hg}$$
• $$P_B^\circ = 600\ \text{mm Hg}$$

A substance with a lower vapour pressure is less volatile because it vaporizes less readily.

Since $$200 < 600$$, liquid A is the least volatile component.

Therefore, the vapour pressure of pure $$B$$ is $$600\ \text{mm Hg}$$, and the least volatile component is A.

Hence, the correct answer is Option (D).

We need to evaluate two statements about the molal depression constant $$K_f$$.

Statement I: Molal depression constant $$K_f$$ is given by $$\frac{M_1 R T_f}{\Delta S_{fus}}$$

The standard expression for the molal freezing point depression constant is:

$$$K_f = \frac{M_1 R T_f^2}{1000 \times \Delta H_{fus}}$$$

where $$M_1$$ is the molar mass of the solvent, $$R$$ is the gas constant, $$T_f$$ is the freezing point of the solvent, and $$\Delta H_{fus}$$ is the enthalpy of fusion.

Since the entropy of fusion is defined as:

$$$\Delta S_{fus} = \frac{\Delta H_{fus}}{T_f}$$$

We can substitute $$\Delta H_{fus} = T_f \times \Delta S_{fus}$$:

$$$K_f = \frac{M_1 R T_f^2}{1000 \times T_f \times \Delta S_{fus}} = \frac{M_1 R T_f}{1000 \times \Delta S_{fus}}$$$

When $$M_1$$ is expressed in kg/mol (i.e., $$M_1$$ already includes the factor of 1000), the expression simplifies to:

$$$K_f = \frac{M_1 R T_f}{\Delta S_{fus}}$$$

Therefore, Statement I is correct.

Statement II: $$K_f$$ for benzene is less than the $$K_f$$ for water.

The known values of $$K_f$$ are:

$$K_f$$ for water $$= 1.86 \text{ K kg mol}^{-1}$$

$$K_f$$ for benzene $$= 5.12 \text{ K kg mol}^{-1}$$

Clearly, $$K_f$$ for benzene $$(5.12)$$ is greater than $$K_f$$ for water $$(1.86)$$.

Therefore, Statement II is incorrect.

Statement I is correct but Statement II is incorrect.

Hence, the correct answer is Option D.

An azeotrope boils at a constant temperature because the vapour phase has the same composition as the liquid phase.
  • If the total vapour pressure of the mixture is higher than that predicted by Raoult’s law (positive deviation), the boiling point of the mixture becomes lower than that of either pure component. Such an azeotrope is called a minimum-boiling azeotrope.
  • If the total vapour pressure is lower than Raoult’s law (negative deviation), the boiling point is higher than that of both components, giving a maximum-boiling azeotrope.

Whether a binary liquid shows positive or negative deviation depends on the relative strength of intermolecular forces:
  • Interactions in the mixture weaker than those within the pure liquids ⇒ molecules escape more easily ⇒ higher vapour pressure ⇒ positive deviation ⇒ minimum-boiling azeotrope.
  • Interactions in the mixture stronger than in the pure liquids (e.g. new hydrogen bonds) ⇒ molecules escape with difficulty ⇒ lower vapour pressure ⇒ negative deviation ⇒ maximum-boiling azeotrope.

Now examine each option.

Case A: $$H_2O + CH_3COOC_2H_5$$ (water + ethyl acetate)
Water is highly polar; ethyl acetate is much less polar and cannot hydrogen-bond effectively with water. Hence, the A-B (unlike) interactions are weaker than A-A and B-B interactions. The mixture shows a large positive deviation. Therefore it forms a minimum-boiling azeotrope (bp ≈ $$70^{\circ}C$$).

Case B: $$C_6H_5OH + C_6H_5NH_2$$ (phenol + aniline)
Both phenol and aniline possess -OH/-NH2 groups that can form strong intermolecular hydrogen bonds with each other. In the mixture the A-B attractions are stronger than those in the pure liquids, so the vapour pressure drops (negative deviation). Consequently, if an azeotrope is formed at all, it will be a maximum-boiling one; it can never be a minimum-boiling azeotrope. Hence this binary system does not show minimum-boiling behaviour.

Case C: $$CS_2 + CH_3COCH_3$$ (carbon disulphide + acetone)
CS2 is non-polar while acetone is polar but cannot hydrogen-bond strongly with CS2. Thus A-B interactions are weaker ⇒ positive deviation ⇒ the pair forms a minimum-boiling azeotrope (bp ≈ $$39^{\circ}C$$).

Case D: $$CH_3OH + CHCl_3$$ (methanol + chloroform)
Though each component can engage in hydrogen bonding, the two form a 1 : 1 complex in the liquid phase which is less stable than the separate hydrogen-bond networks in the pure liquids. Hence the net A-B interaction is weaker, giving a positive deviation and a minimum-boiling azeotrope (bp ≈ $$55.9^{\circ}C$$ at 1 atm).

Only the mixture in Option B fails to exhibit minimum-boiling behaviour.

Correct answer: Option B ($$C_6H_5OH + C_6H_5NH_2$$)

For an ideal liquid mixture Raoult’s law is valid.
If $$P_A^0$$ and $$P_B^0$$ are the vapour-pressures of the pure liquids, then the partial pressures over the solution are

$$p_A = x_A P_A^0$$ and $$p_B = x_B P_B^0$$ $$-(1)$$

The total pressure is $$P = p_A + p_B$$. The mole fractions of the components in the vapour phase are defined as

$$y_A = \frac{p_A}{P}$$ and $$y_B = \frac{p_B}{P}$$ $$-(2)$$

Divide the two expressions in $$(2)$$:

$$\frac{y_A}{y_B} = \frac{p_A/P}{p_B/P} = \frac{p_A}{p_B}$$

Now use $$(1)$$:

$$\frac{y_A}{y_B} = \frac{x_A P_A^0}{x_B P_B^0} = \frac{x_A}{x_B}\,\frac{P_A^0}{P_B^0}$$ $$-(3)$$

The data given are $$P_A^0 = 350\ \text{mm Hg}$$ and $$P_B^0 = 750\ \text{mm Hg}$$, so

$$\frac{P_A^0}{P_B^0} = \frac{350}{750} = 0.467 \lt 1$$

Substituting into $$(3)$$:

$$\frac{y_A}{y_B} = \frac{x_A}{x_B}\,(0.467)$$

Because the multiplier $$0.467$$ is less than $$1$$, we obtain

$$\frac{y_A}{y_B} \lt \frac{x_A}{x_B} \quad\Longrightarrow\quad \frac{x_A}{x_B} \gt \frac{y_A}{y_B}$$

Hence the correct relation is given in Option C.

Answer - Option C

First, list the given data for the two systems.

Solution 1 (stock solution): 10 mol solute x + 10 L water.
Hence, the concentration is $$\frac{10\;\text{mol}}{10\;\text{L}} = 1\;\text{M}$$.

Solution 2 is prepared by mixing 1 L of Solution 1 with 1 mol of solute x and 1 L of water.
• Solute coming from 1 L of Solution 1 = 1 mol.
• Extra solute added = 1 mol.
Total solute in Solution 2 = 1 mol + 1 mol = 2 mol.
• Volume coming from 1 L of Solution 1 = 1 L.
• Extra water added = 1 L.
Total volume of Solution 2 = 1 L + 1 L = 2 L.

Therefore, the concentration of Solution 2 is $$\frac{2\;\text{mol}}{2\;\text{L}} = 1\;\text{M}$$, exactly the same as Solution 1.

Now analyse each property.

Case A: Molar heat capacity (heat capacity per mole of solution) is an intensive property; it depends only on composition, not on the amount. Because the mole-ratio of solute to solvent remains 1 M for both solutions, the molar heat capacity stays unchanged.

Case B: Density is also an intensive property (mass per unit volume). Since the composition is identical in both solutions, the density remains the same.

Case C: Concentration has already been shown to stay at 1 M. Hence, no change occurs.

Case D: Gibbs free energy $$G$$ is an extensive property; it is directly proportional to the amount of matter present. Solution 2 contains twice the amount of both solute and solvent compared with the 1 L portion of Solution 1, so its total Gibbs free energy is different (specifically, doubled) from that of Solution 1.

Therefore, the only property that changes when the system passes from Solution 1 to Solution 2 is the Gibbs free energy.

Correct choice: Option D (Gibbs free energy).

Molarity is given by

$$M=\frac{n}{V}$$

In this problem, a fixed mass of NaCl is added, so the number of moles of solute, $$n$$, remains constant. Since the beaker is closed, there is no loss of water due to evaporation. Therefore, only the volume of the solution changes with temperature.

Hence,

$$M\propto \frac{1}{V}$$

Thus, molarity varies inversely with the volume of the solution.

Water exhibits anomalous expansion.

• From $$0^\circ\text{C}$$ to $$4^\circ\text{C}$$, water contracts on heating, so its volume decreases.
• At $$4^\circ\text{C}$$, water attains minimum volume and maximum density.
• Above $$4^\circ\text{C}$$, water expands on heating, so its volume increases.

Since the solution is dilute, it is assumed to follow the same trend as water.

Therefore,

• From $$1^\circ\text{C}$$ to $$4^\circ\text{C}$$, volume decreases, so molarity increases.
• At $$4^\circ\text{C}$$, volume is minimum, so molarity is maximum.
• From $$4^\circ\text{C}$$ to $$25^\circ\text{C}$$, volume increases, so molarity decreases.

Hence, the graph of molarity versus temperature first increases, reaches a maximum at approximately $$4^\circ\text{C}$$, and then decreases.

Therefore, the correct graph is Option (B).

For each pair we first decide whether the solution shows ideal behaviour, positive deviation or negative deviation from Raoult’s law. Then we match the corresponding properties from List-II.

Case A: Chloroform (CHCl₃) + Acetone (CH₃COCH₃)
• CHCl₃ has an acidic hydrogen that forms strong hydrogen bonds with the carbonyl oxygen of acetone.
• The intermolecular attraction between unlike molecules becomes stronger than that between like molecules.
• Stronger attractions ↓ total vapour pressure → these mixtures show negative deviation from Raoult’s law.
• Negative deviation gives a solution whose boiling point is higher than either pure component, i.e. a maximum-boiling azeotrope.
Thus (A) → (III).

Case B: Ethanol (C₂H₅OH) + Water (H₂O)
• When ethanol mixes with water, some hydrogen bonds of pure water are broken; the unlike interactions are weaker than the strong H-bond network of water.
• Weaker attractions ↑ total vapour pressure → positive deviation from Raoult’s law.
• Positive deviation produces a minimum-boiling azeotrope.
Thus (B) → (I).

Case C: Benzene (C₆H₆) + Toluene (C₆H₅CH₃)
• Both are non-polar aromatic liquids with very similar molecular sizes and intermolecular forces.
• Their mixture obeys Raoult’s law almost exactly: $$\Delta H_{mix} = 0$$ and $$\Delta V_{mix} = 0$$.
Thus (C) → (IV).

Case D: Acetic acid (CH₃COOH) in Benzene (C₆H₆)
• In a non-polar solvent like benzene, acetic acid molecules associate through hydrogen bonding to form dimers: $$2\,CH_3COOH \rightleftharpoons (CH_3COOH)_2$$.
Thus (D) → (II).

Collecting the matches:
(A)-(III), (B)-(I), (C)-(IV), (D)-(II).

The option having this sequence is Option A.

Final answer: Option A.

A living cell contains $$0.9\%$$ (w/w) glucose solution and is immersed in a solution with equal mole fractions of glucose and water.

Equal mole fractions of glucose and water means $$x_{glucose} = x_{water} = 0.5$$, so the moles of glucose and water are equal. Assuming 1 mol of each, the mass of glucose is $$180\text{ g}$$ and the mass of water is $$18\text{ g}$$, giving a total mass of $$198\text{ g}$$ and a percentage by mass of $$\%(w/w) = \frac{180}{198}\times 100 \approx 90.9\%$$ for the external solution.

The cell has $$0.9\%$$ glucose, while the external solution has approximately $$90.9\%$$ glucose, making the external solution extremely hypertonic compared to the cell.

Because the external solution has a much higher solute concentration and a correspondingly lower water activity, water will flow out of the cell by osmosis, causing the cell to shrink dramatically. However, the calculated concentration (~90.9\%) does not match any of the specific percentages given in options A, B, or C.

The correct answer is Option D) None of these.

For a binary solution of two volatile liquids, find the slope and intercept of $$\frac{1}{x_1}$$ vs $$\frac{1}{y_1}$$.

Apply Raoult's law

Total pressure: $$P = x_1P_1° + x_2P_2° = x_1P_1° + (1-x_1)P_2° = P_2° + x_1(P_1° - P_2°)$$

Mole fraction in vapor: $$y_1 = \frac{x_1P_1°}{P} = \frac{x_1P_1°}{P_2° + x_1(P_1° - P_2°)}$$

Take reciprocal

$$\frac{1}{y_1} = \frac{P_2° + x_1(P_1° - P_2°)}{x_1P_1°} = \frac{P_2°}{x_1P_1°} + \frac{P_1° - P_2°}{P_1°}$$

$$\frac{1}{y_1} = \frac{P_2°}{P_1°} \cdot \frac{1}{x_1} + \frac{P_1° - P_2°}{P_1°}$$

Compare with y = mx + c format

Plotting $$\frac{1}{y_1}$$ (y-axis) vs $$\frac{1}{x_1}$$ (x-axis) -- but the question asks for $$\frac{1}{x_1}$$ vs $$\frac{1}{y_1}$$. Let me rearrange:

$$\frac{1}{x_1} = \frac{P_1°}{P_2°} \cdot \frac{1}{y_1} - \frac{P_1° - P_2°}{P_2°}$$

$$= \frac{P_1°}{P_2°} \cdot \frac{1}{y_1} + \frac{P_2° - P_1°}{P_2°}$$

Slope = $$\frac{P_1°}{P_2°}$$, Intercept = $$\frac{P_2° - P_1°}{P_2°}$$

The correct answer is Option 2: $$\frac{P_1°}{P_2°}$$ and $$\frac{P_2° - P_1°}{P_2°}$$.

Using $$\Delta T_b = i \cdot K_b \cdot m$$ with $$K_b = 0.52$$ K kg mol$$^{-1}$$.

For $$AX_2$$: $$\Delta T_b = 0.0156°C$$, molality = $$0.01$$ mol/kg.

$$i = \frac{0.0156}{0.52 \times 0.01} = 3$$

Since $$AX_2 \to A^{2+} + 2X^-$$ gives 3 particles, $$AX_2$$ is fully ionised.

For $$AY_2$$: $$\Delta T_b = 0.0260°C$$, molality = $$\frac{25.4/250}{2} = 0.0508$$ mol/kg.

$$i = \frac{0.0260}{0.52 \times 0.0508} \approx 1$$

Since $$i \approx 1$$, $$AY_2$$ is completely unionised.

The correct answer is Option 1: $$AX_2$$ is fully ionised while $$AY_2$$ is completely unionised.

The normal boiling point of pure water is $$373 \text{ K}$$.

Total moles of solute:
Ethylene glycol $$= 2 \text{ mol}$$
Glucose $$= 2 \text{ mol}$$
Therefore, $$n_{\text{total}} = 2 + 2 = 4 \text{ mol}$$.

Mass of solvent (water) $$= 500 \text{ g} = 0.5 \text{ kg}$$.

Molality is defined as $$m = \frac{n_{\text{solute}}}{\text{mass of solvent in kg}}$$. Hence
$$m = \frac{4}{0.5} = 8 \text{ mol kg}^{-1}$$ $$-(1)$$

The elevation in boiling point is given by the formula
$$\Delta T_b = i \, K_b \, m$$ $$-(2)$$
For non-electrolytes like ethylene glycol and glucose, the van’t Hoff factor $$i = 1$$.
Given $$K_b = 0.52 \text{ K kg mol}^{-1}$$.

Substituting values into $$(2)$$:
$$\Delta T_b = 1 \times 0.52 \times 8 = 4.16 \text{ K}$$.

Boiling point of the solution:
$$T_{\text{boil}} = 373 \text{ K} + 4.16 \text{ K} = 377.16 \text{ K}$$.

Rounded to one decimal place, $$T_{\text{boil}} \approx 377.3 \text{ K}$$.

Hence, the boiling point of the solution is $$377.3 \text{ K}$$ (Option B).

We need to find the freezing point depression constant ($$K_f$$) of a solvent.

Mass of solvent = 50 g = 0.050 kg

Mass of non-volatile solute = 1 g

Molar mass of solute = 256 g/mol

Decrease in freezing point $$\Delta T_f = 0.40$$ K

$$\Delta T_f = K_f \times m$$

where $$m$$ is the molality of the solution.

Moles of solute = $$\frac{1}{256}$$ mol = 0.003906 mol

Molality $$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{1/256}{0.050} = \frac{1}{12.8} = 0.078125$$ mol/kg

$$K_f = \frac{\Delta T_f}{m} = \frac{0.40}{0.078125} = 5.12$$ K kg mol$$^{-1}$$

The correct answer is Option 4: $$5.12$$ K kg mol$$^{-1}$$.

Arrange the solutions in order of increasing boiling points.

We recall that boiling point elevation depends on the total solute particle concentration.

$$\Delta T_b = i \cdot K_b \cdot m$$

Here, $$i$$ is the van't Hoff factor.

Next, we calculate the effective concentration for each solution.

For (i) $$10^{-4}$$ M NaCl, which dissociates as NaCl → Na⁺ + Cl⁻, the van't Hoff factor is $$i = 2$$ giving an effective concentration of $$2 \times 10^{-4}$$ M.

For (ii) $$10^{-4}$$ M urea, a non-electrolyte with $$i = 1$$, the effective concentration is $$10^{-4}$$ M.

For (iii) $$10^{-3}$$ M NaCl with $$i = 2$$, the effective concentration becomes $$2 \times 10^{-3}$$ M.

For (iv) $$10^{-2}$$ M NaCl with $$i = 2$$, the effective concentration becomes $$2 \times 10^{-2}$$ M.

Ordering these effective concentrations yields

$$10^{-4} < 2 \times 10^{-4} < 2 \times 10^{-3} < 2 \times 10^{-2}$$

Hence, in terms of the given options:

$$(\text{ii}) < (\text{i}) < (\text{iii}) < (\text{iv})$$

Therefore, the solutions in order of increasing boiling point are (ii) < (i) < (iii) < (iv).

1. Vapor Pressure Lowering:

   The addition of a non-volatile solute lowers the vapor pressure of the liquid solvent. Therefore, at any given temperature, the vapor pressure curve of the solution always lies below the vapor pressure curve of the pure liquid solvent.

2. Definition of Freezing Point:

   The freezing point is the specific temperature at which the liquid phase and the solid phase of a substance coexist in dynamic equilibrium, meaning they have the exact same vapor pressure.

3. Intersection Points on the Graph:

   • The freezing point of the pure solvent ($$T_f^0$$) is determined by the intersection of the liquid solvent curve with the frozen (solid) solvent curve.
   • The freezing point of the solution ($$T_f$$) is determined by the intersection of the solution curve with the frozen (solid) solvent curve.

Deducing the Correct Graph:

Because the solution curve lies below the pure liquid solvent curve, its intersection with the steep frozen solvent line occurs at a lower temperature compared to the pure liquid solvent:

$$T_f < T_f^0$$

The depression in freezing point ($$\Delta T_f$$) is the horizontal distance between these two temperatures on the temperature ($$T/K$$) axis:

$$\Delta T_f = T_f^0 - T_f$$

The correct graph must clearly show three distinct lines: the pure liquid solvent (top curve), the solution (bottom curve), and the frozen solvent (steep intersecting line), with $$\Delta T_f$$ representing the drop in freezing temperature along the x-axis.

Vapor pressure decreases by 10 mm Hg when mole fraction of solute is 0.2. Find mole fraction of solvent when decrease is 20 mm Hg.

Apply Raoult's law for relative lowering

$$\frac{\Delta P}{P°} = x_{solute}$$

$$\frac{10}{P°} = 0.2 \Rightarrow P° = 50$$ mm Hg

For decrease of 20 mm Hg

$$x_{solute} = \frac{20}{50} = 0.4$$

$$x_{solvent} = 1 - 0.4 = 0.6$$

The correct answer is Option 4: 0.6.

For dissolved oxygen we must convert ppm (mass ratio) to molality (moles of $$\mathrm{O_2}$$ per kg of water).

Case 1: Work with exactly $$1\ \text{L}$$ of sea water (any convenient volume can be chosen).

Density is $$2\ \text{g mL}^{-1}$$, so mass of this $$1\ \text{L} = 1000\ \text{mL}$$ of solution is
$$1000 \times 2 = 2000\ \text{g}$$.

Case 2: Separate solute (NaCl) from solvent (water).

The solution is $$6\ \text{M}$$ in NaCl, i.e. $$6$$ moles NaCl per litre.
Molar mass of NaCl is $$58.5\ \text{g mol}^{-1}$$, therefore mass of NaCl in $$1\ \text{L}$$ is
$$6 \times 58.5 = 351\ \text{g}$$.

Mass of water (solvent) present = total mass − mass of NaCl:
$$2000 - 351 = 1649\ \text{g} = 1.649\ \text{kg}$$.

Case 3: Find moles of dissolved $$\mathbf{O_2}$$ present in the same litre.

Concentration of $$\mathrm{O_2}$$ is $$5.8\ \text{ppm}$$. For mass ratios, $$1\ \text{ppm} = 1\ \text{mg}$$ solute per $$1\ \text{kg}$$ solution.

Hence in $$1\ \text{kg}$$ solution, mass of $$\mathrm{O_2} = 5.8\ \text{mg} = 0.0058\ \text{g}$$.
Our sample is $$2000\ \text{g} = 2\ \text{kg}$$ of solution, so

Mass of $$\mathrm{O_2} = 2 \times 0.0058 = 0.0116\ \text{g}$$.

Molar mass of $$\mathrm{O_2} = 32\ \text{g mol}^{-1}$$, therefore moles of $$\mathrm{O_2}$$ are
$$\frac{0.0116}{32} = 3.625 \times 10^{-4}\ \text{mol}$$.

Case 4: Convert to molality.

Molality $$m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}$$ = $$\frac{3.625 \times 10^{-4}}{1.649} = 2.20 \times 10^{-4}\ m$$.

Expressed as $$x \times 10^{-4}\ m$$, the value of $$x \approx 2.20$$, which to the nearest integer is $$2$$.

Therefore the concentration of dissolved oxygen is $$\mathbf{2 \times 10^{-4}\ m}$$ (nearest integer $$x = 2$$).

The precipitation reaction involved is:
NaI + AgNO3 → AgI ↓ + NaNO3

The stoichiometry is 1 : 1, so the moles of NaI present in the sample equal the moles of AgI obtained.

Molar mass of silver iodide (AgI):
Ag = 108 g mol$$^{-1}$$, I = 127 g mol$$^{-1}$$
Therefore, molar mass of AgI = $$108 + 127 = 235$$ g mol$$^{-1}$$.

Moles of AgI formed:
$$n = \frac{4.74}{235} = 0.0202$$ mol

Because of the 1 : 1 ratio, moles of NaI in 20 mL solution = $$0.0202$$ mol.

Volume of the NaI solution = 20 mL = 0.020 L.

Molarity, $$M = \frac{\text{moles}}{\text{volume in L}} = \frac{0.0202}{0.020} = 1.01$$ M.

Rounded to two significant figures, the molarity is approximately $$1$$ M.

Answer : 1 M

Let the salt be written as $$MX_3$$. In water it dissociates as

$$MX_3 \;\rightarrow\; M^{3+} + 3\,X^-$$

Thus one formula unit produces $$n = 1 + 3 = 4$$ particles when the dissociation is complete.

For any electrolyte, the van’t Hoff factor $$i$$ is related to the degree of dissociation $$\alpha$$ by

$$i = 1 + (n-1)\,\alpha \quad -(1)$$

In the present case, $$n = 4$$, and the experimental value is $$i = 2$$. Substitute these values in $$(1)$$:

$$2 = 1 + (4-1)\,\alpha$$

$$2 = 1 + 3\,\alpha$$

$$3\,\alpha = 1$$

$$\alpha = \frac{1}{3} = 0.333...$$

The percentage dissociation is therefore

$$\alpha \times 100\% = 0.333... \times 100\% \approx 33\%$$

Nearest integer = 33 %.

The boiling-point elevation of a dilute solution is given by
$$\Delta T_b = K_b \, m$$
where $$K_b$$ is the molal elevation constant and $$m$$ is the molality of the solution.

Case 1: 1 g of $$AB$$ in 15 g (0.015 kg) of water raises the boiling point by 2.7 K.
For water, $$K_b = 0.5 \text{ K kg mol}^{-1}$$, so

$$m_{\!AB} = \frac{\Delta T_b}{K_b} = \frac{2.7}{0.5}=5.4 \text{ mol kg}^{-1}$$

Molality is also $$m = \dfrac{w/M}{W}$$, hence
$$5.4 = \frac{1/M_{AB}}{0.015} \;\;\Rightarrow\;\; M_{AB} = 12.34 \text{ g mol}^{-1}$$

Case 2: 1 g of $$AB_2$$ in the same quantity of water raises the boiling point by 1.5 K.

$$m_{AB_2} = \frac{1.5}{0.5}=3.0 \text{ mol kg}^{-1}$$

$$3.0 = \frac{1/M_{AB_2}}{0.015} \;\;\Rightarrow\;\; M_{AB_2} = 22.22 \text{ g mol}^{-1}$$

Let the atomic masses be $$M_A$$ and $$M_B$$. Then

$$M_A + M_B = 12.34 \;\;-(1)$$
$$M_A + 2M_B = 22.22 \;\;-(2)$$

Subtract $$(1)$$ from $$(2)$$:

$$M_B = 22.22 - 12.34 = 9.88 \text{ g mol}^{-1}$$

Substitute in $$(1)$$ to find $$M_A$$:

$$M_A = 12.34 - 9.88 = 2.46 \text{ g mol}^{-1}$$

The atomic mass of A is therefore $$2.46 \text{ amu} \approx 2.5 \text{ amu}$$.
Expressed as $$\times 10^{-1}$$, this is
$$2.5 \text{ amu} = 25 \times 10^{-1} \text{ amu}$$

Hence the required nearest integer is 25.

If $$A_2B$$ is 30% ionised in an aqueous solution, find the van't Hoff factor (i) expressed as $$\_\_\_ \times 10^{-1}$$.

The dissociation equation is $$A_2B \rightarrow 2A^+ + B^{2-}$$, and one formula unit produces 3 ions (2 cations + 1 anion), so $$n = 3$$.

The van't Hoff factor formula is $$i = 1 + \alpha(n - 1)$$, where $$\alpha = 0.30$$ and $$n = 3$$. Thus, $$i = 1 + 0.30(3 - 1) = 1 + 0.60 = 1.60$$.

Expressing this in the required form gives $$i = 1.60 = 16 \times 10^{-1}$$, so the answer is 16.

The mole fraction of a component is defined as the ratio of the number of moles of that component to the total number of moles present in the mixture.

For a mixture containing $$n_A$$ moles of A, $$n_B$$ moles of B, and $$n_C$$ moles of C, the total number of moles is, $$n_A+n_B+n_C$$

The mole fraction of component C is obtained by dividing the number of moles of C by the total number of moles in the mixture.

Therefore, $$\chi_C=\frac{n_C}{n_A+n_B+n_C}$$

Hence, the correct answer is, $$\frac{n_C}{n_A+n_B+n_C}$$

Among the given quantities:

- Molarity depends on volume of solution, which changes with temperature. So molarity changes with temperature.

- Mass percentage depends only on masses, independent of temperature.

- Molality depends on mass of solvent, independent of temperature.

- Mole fraction depends on number of moles, independent of temperature.

The answer is Molarity, which corresponds to Option (1).

We need the solution with highest depression in freezing point. $$\Delta T_f = iK_fm$$ where $$i$$ is van't Hoff factor and $$m$$ is molality.

Option 1: 180 g acetic acid (M=60) in 1 L water. Moles = 3. $$m \approx 3$$. Acetic acid dissociates slightly in water, $$i > 1$$. $$\Delta T_f \approx 3 \times 1.86 \times i > 5.58$$.

Option 2: 180 g acetic acid in benzene. Acetic acid dimerizes in benzene, $$i \approx 0.5$$. $$\Delta T_f$$ much lower.

Option 3: 180 g benzoic acid (M=122) in benzene. Moles $$\approx 1.48$$. Benzoic acid also dimerizes in benzene, $$i \approx 0.5$$.

Option 4: 180 g glucose (M=180) in water. Moles = 1. Non-electrolyte, $$i = 1$$. $$\Delta T_f = 1.86$$.

Option 1 has the highest molality (3 mol/kg) and in water, acetic acid ionizes ($$i > 1$$), giving the largest freezing point depression.

The correct answer is Option 1.

We need to identify the mixture that shows positive deviation from Raoult's Law.

Key Concept: Positive deviation from Raoult's Law occurs when the interactions between unlike molecules (A-B) are weaker than the interactions between like molecules (A-A and B-B). This causes the vapor pressure of the solution to be higher than predicted by Raoult's Law, because molecules escape the solution more easily.

Option 1: Acetone + Aniline ($$(CH_3)_2CO + C_6H_5NH_2$$):

Acetone and aniline form hydrogen bonds with each other (C=O of acetone with N-H of aniline). A-B interactions are stronger than average A-A and B-B interactions. This gives negative deviation.

Option 2: Chloroform + Benzene ($$CHCl_3 + C_6H_6$$):

Chloroform and benzene have weak attractive interactions (dipole-induced dipole). These are comparable to the original interactions, giving nearly ideal behavior or slight negative deviation.

Option 3: Chloroform + Acetone ($$CHCl_3 + (CH_3)_2CO$$):

Chloroform forms hydrogen bonds with acetone (C-H...O=C). The A-B interactions are stronger, giving negative deviation.

Option 4: Acetone + Carbon disulfide ($$(CH_3)_2CO + CS_2$$):

Acetone has dipole-dipole interactions (polar molecule), while $$CS_2$$ is non-polar. When mixed, the strong dipole-dipole interactions of acetone are disrupted, and the new A-B interactions (dipole-induced dipole) are weaker. This makes it easier for molecules to escape into the vapor phase, resulting in positive deviation.

The correct answer is Option (4): $$(CH_3)_2CO + CS_2$$.

We need to determine what happens to the freezing point of benzene when a small quantity of naphthalene is added.

Key Concept: Depression in Freezing Point (Colligative Property) When a non-volatile solute is dissolved in a solvent, the freezing point of the solution decreases. This is known as the depression of freezing point, a colligative property.

The depression in freezing point is given by:

$$\Delta T_f = K_f \cdot m$$

where $$K_f$$ is the cryoscopic constant of the solvent and $$m$$ is the molality of the solution.

Naphthalene is a non-volatile, non-electrolyte solute. When it is dissolved in benzene (the solvent), it forms a solution and lowers the vapour pressure of benzene. This causes the solid-liquid equilibrium to shift, requiring a lower temperature for freezing to occur. Since $$\Delta T_f = K_f \cdot m > 0$$, the freezing point of the solution is lower than that of pure benzene.

Therefore, the freezing point of benzene decreases when naphthalene is added.

Answer: Option 4 - Decreases

A solution showing negative deviation from Raoult's law has stronger intermolecular forces between unlike molecules than like molecules.

This results in:

- Decreased vapour pressure (molecules are held more tightly in solution, fewer escape to vapor phase)

- Increased boiling point (higher temperature needed to reach atmospheric pressure since vapour pressure is lower)

The answer is: decreased vapour pressure, increased boiling point, which corresponds to Option (4).

We need to find the concentration of intermediate $$B$$ in the consecutive reaction $$A \xrightarrow{K_1} B \xrightarrow{K_2} C$$, using the steady-state approximation.

The intermediate $$B$$ is formed from $$A$$ (with rate constant $$K_1$$) and is consumed to form $$C$$ (with rate constant $$K_2$$). The rate of change of $$[B]$$ is:

$$\frac{d[B]}{dt} = K_1[A] - K_2[B]$$

The first term represents formation of $$B$$ from $$A$$, and the second term represents consumption of $$B$$ to form $$C$$.

The steady-state approximation assumes that after an initial transient period, the concentration of the intermediate $$B$$ remains approximately constant. This means:

$$\frac{d[B]}{dt} = 0$$

Setting the rate equation to zero:

$$K_1[A] - K_2[B] = 0$$

$$K_2[B] = K_1[A]$$

$$[B] = \frac{K_1}{K_2}[A]$$

The correct answer is Option (2): $$(K_1/K_2)[A]$$.

Total surface area available for adsorption on $$1 \text{ g}$$ of charcoal is given to be
$$A = 1.5 \times 10^{2}\; \text{m}^{2}$$

Step 1: Moles of acetic acid taken
Solution volume = $$100\;\text{mL} = 0.1\;\text{L}$$
Molarity = $$0.5\;\text{M}$$
$$n_{\text{taken}} = 0.1 \times 0.5 = 0.05\;\text{mol}$$

Step 2: Moles of acetic acid remaining unadsorbed
Volume of NaOH used = $$40\;\text{mL} = 0.04\;\text{L}$$
Molarity of NaOH = $$1\;\text{M}$$
For $$\text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}$$ the stoichiometry is 1 : 1, so
$$n_{\text{unadsorbed}} = 0.04 \times 1 = 0.04\;\text{mol}$$

Step 3: Moles of acetic acid adsorbed on charcoal
$$n_{\text{adsorbed}} = n_{\text{taken}} - n_{\text{unadsorbed}} = 0.05 - 0.04 = 0.01\;\text{mol}$$

Step 4: Number of molecules adsorbed
Avogadro number $$N_A = 6.0 \times 10^{23}\;\text{mol}^{-1}$$
$$N = 0.01 \times 6.0 \times 10^{23} = 6.0 \times 10^{21}$$ molecules

Step 5: Surface area occupied by one molecule
$$\text{Area per molecule} = \frac{A}{N} = \frac{1.5 \times 10^{2}}{6.0 \times 10^{21}} = 0.25 \times 10^{-19}\;\text{m}^{2} = 2.5 \times 10^{-20}\;\text{m}^{2}$$

Step 6: Expressing in the required form
$$2.5 \times 10^{-20}\;\text{m}^{2} = 2.5 \times 10^{3} \times 10^{-23}\;\text{m}^{2}$$

Hence $$P = 2500$$.

Final answer: 2500

Iron(III) catalyses the reaction between iodide and persulphate ions. Identify the correct mechanism.

The overall reaction is $$2I^- + S_2O_8^{2-} \rightarrow I_2 + 2SO_4^{2-}$$.

In the catalytic cycle of Fe³⁺/Fe²⁺, Fe³⁺ oxidizes I⁻: $$2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2$$. Fe²⁺ is then re-oxidized by persulphate (Fe²⁺ reduces persulphate): $$2Fe^{2+} + S_2O_8^{2-} \rightarrow 2Fe^{3+} + 2SO_4^{2-}$$.

So: A (Fe³⁺ oxidises iodide) and D (Fe²⁺ reduces persulphate) are correct.

The correct answer is Option (4): A and D only.

The elevation of the boiling point for a dilute solution is given by the relation
$$\Delta T_b = i\,K_b\,m$$
where $$i$$ is the van’t Hoff factor, $$K_b$$ is the ebullioscopic constant of the solvent (same for both vessels because the solvent is water and the temperature is identical), and $$m$$ is the molality of the solute.

Because both vessels contain the same solvent (water) at the same temperature, $$K_b$$ is identical for the two solutions. Hence the ratio of the boiling-point elevations equals the product of the ratios of the van’t Hoff factors and the molalities:

$$\frac{\Delta T_{b1}}{\Delta T_{b2}} \;=\; \frac{i_X}{i_Y}\;\times\;\frac{m_X}{m_Y}$$

Step 1: Ratio of van’t Hoff factors
Given that the van’t Hoff factor for solute X is 1.2 times that for solute Y,
$$\frac{i_X}{i_Y} = 1.2$$

Step 2: Ratio of molalities
Molality is defined as $$m = \dfrac{\text{moles of solute}}{\text{kilograms of solvent}}$$.
With $$w_2$$ g of solute dissolved in $$w_1$$ g of water, the molality for any solute is
$$m = \frac{1000\,w_2}{M\,w_1}$$
where $$M$$ is the molar mass of the solute (in g mol$$^{-1}$$).

The two vessels have the same $$w_1$$ and $$w_2$$, so the ratio of their molalities depends only on their molar masses:
$$\frac{m_X}{m_Y} = \frac{M_Y}{M_X}$$

We are told that the molar mass of X is 80 % of that of Y:
$$M_X = 0.8\,M_Y \;\;\Longrightarrow\;\; \frac{M_Y}{M_X} = \frac{1}{0.8} = 1.25$$
Therefore,
$$\frac{m_X}{m_Y} = 1.25$$

Step 3: Ratio of boiling-point elevations
Combining the two ratios:
$$\frac{\Delta T_{b1}}{\Delta T_{b2}} = 1.2 \times 1.25 = 1.50$$

Thus, the elevation of the boiling point in Vessel 1 is 1.50 times (or 150 %) that in Vessel 2.

Answer: 150 %

The van't Hoff factor, denoted as $$i$$, measures the extent of dissociation or association of a solute in solution. For electrolytes, it is given by:

$$i = \frac{\text{observed colligative property}}{\text{calculated colligative property assuming no dissociation}}$$

NaCl is a strong electrolyte that dissociates completely in ideal conditions:

$$\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-$$

The theoretical van't Hoff factor for complete dissociation is 2, as one formula unit produces two ions.

However, in real solutions, interionic attractions reduce the effective number of ions, making the observed $$i$$ less than 2. This effect is more pronounced at higher concentrations due to stronger interionic interactions.

Given the concentrations:

• Solution A: 0.1 M (highest concentration)
• Solution B: 0.01 M
• Solution C: 0.001 M (lowest concentration)

As concentration decreases, interionic attractions weaken, and the van't Hoff factor increases towards the theoretical value of 2.

Therefore:

• For A (0.1 M), $$i_A$$ is the smallest due to strongest interionic attractions.
• For B (0.01 M), $$i_B$$ is greater than $$i_A$$ but less than $$i_C$$.
• For C (0.001 M), $$i_C$$ is the largest and closest to 2.

Thus, the order is $$i_A < i_B < i_C$$.

Comparing with the options:

• A. $$i_A < i_B < i_C$$ (matches)
• B. $$i_A < i_C < i_B$$ (incorrect)
• C. $$i_A = i_B = i_C$$ (incorrect, as concentration affects i)
• D. $$i_A > i_B > i_C$$ (incorrect, opposite order)

The correct answer is option A.

Δ$$T_f=K_f$$ × m . 

Solvent is water (2.7kg). 

Solute is acetic acid: 2700/60=45 mol. 

m=$$\frac{\ 45}{2.7}$$=16.67.

Δ$$T_f=$$1.86×16.67=31°C. 

Freezing point = 0-31=-31°C.

The answer is $$\boxed{31}$$.

Find the mass of ethylene glycol needed to protect 18.6 kg water from freezing at $$-24°C$$.

$$ \Delta T_f = K_f \times m $$

where $$\Delta T_f$$ is the depression in freezing point, $$K_f$$ is the molal freezing point depression constant, and $$m$$ is molality.

$$\Delta T_f = 0 - (-24) = 24$$ K.

$$ m = \frac{\Delta T_f}{K_f} = \frac{24}{1.86} = 12.903 \text{ mol/kg} $$

$$ \text{Moles} = m \times \text{mass of solvent (kg)} = 12.903 \times 18.6 = 240 \text{ mol} $$

Molar mass of ethylene glycol = 62 g/mol.

$$ \text{Mass} = 240 \times 62 = 14880 \text{ g} = 14.88 \text{ kg} \approx 15 \text{ kg} $$

The answer is 15 kg.

The solute is non-volatile and non-electrolytic, hence the boiling-point elevation formula is

$$\Delta T_b = K_b\,m \quad -(1)$$
where $$m$$ is the molality of the solution.

Given $$\Delta T_b = 2^{\circ}\text{C}$$ and $$K_b = 0.52\ \text{K·kg mol}^{-1}$$, from $$(1)$$

$$m = \frac{\Delta T_b}{K_b} = \frac{2}{0.52} \approx 3.85\ \text{mol kg}^{-1} \quad -(2)$$

The mass of water (solvent) is 100 g = 0.100 kg.
Hence the number of moles of solute is

$$n_{\text{solute}} = m \times \text{mass of solvent (kg)}$$ $$= 3.85 \times 0.100 \approx 0.385\ \text{mol} \quad -(3)$$

Moles of water

$$n_{\text{water}} = \frac{100\ \text{g}}{18\ \text{g mol}^{-1}} \approx 5.56\ \text{mol} \quad -(4)$$

Because the solute concentration is very small compared with that of the solvent, the mole fraction of the solute can be approximated by

$$x_{\text{solute}} \approx \frac{n_{\text{solute}}}{n_{\text{water}}}$$

Substituting $$(3)$$ and $$(4)$$:

$$x_{\text{solute}} \approx \frac{0.385}{5.56} \approx 0.069 \quad -(5)$$

For a non-volatile solute, the relative lowering of vapour pressure is

$$\frac{P^\circ - P}{P^\circ} = x_{\text{solute}} \quad -(6)$$

where $$P^\circ$$ is the vapour pressure of pure solvent and $$P$$ is that of the solution.

At the normal boiling point of water (100 °C), $$P^\circ = 1\ \text{atm} = 760\ \text{mm Hg}$$. Using $$(5)$$ in $$(6)$$:

$$\frac{760 - P}{760} = 0.069$$ $$P = 760\,(1 - 0.069) \approx 760 \times 0.931 \approx 707\ \text{mm Hg}$$

Nearest integer:

Vapour pressure of the solution = $$707\ \text{mm Hg}$$.

31.4% H₂SO₄ by mass, density 1.25 g/mL.

In 1 L: mass = 1250 g. Mass of H₂SO₄ = 0.314 × 1250 = 392.5 g.

Moles = 392.5/98 = 4.005 ≈ 4 M. The answer is $$\boxed{4}$$.

Find the overall order of the reaction $$A + B \rightarrow C$$.

Since the time for A to become $$\tfrac{1}{4}$$ of its initial concentration is twice the time required to reach half its initial concentration, and because for a first-order reaction $$t_{1/4} = \frac{2\ln 2}{k} = 2 \times \frac{\ln 2}{k} = 2t_{1/2}\,$$, the reaction exhibits first-order kinetics with respect to A.

Next, the plot of the concentration of B versus time is a straight line with a negative slope and a positive intercept, which corresponds to the integrated rate law $$[B] = [B]_0 - kt$$ for a zero-order reaction in B.

Using these results, the overall order of the reaction is $$1\;( \text{for A} ) + 0\;( \text{for B} ) = 1\,$$.

The correct answer is $$1$$.

For the reaction $$2A + B \rightarrow C + D$$, let $$\text{Rate} = k[A]^a[B]^b$$.

From experiments I and IV (where B is constant at 0.1 M):

$$ \frac{r_{IV}}{r_I} = \left(\frac{0.4}{0.1}\right)^a = 4^a = \frac{2.40 \times 10^{-2}}{6.0 \times 10^{-3}} = 4 $$
$$ 4^a = 4 \Rightarrow a = 1 $$

From experiments II and III (where A is constant at 0.3 M):

$$ \frac{r_{III}}{r_{II}} = \left(\frac{0.4}{0.2}\right)^b = 2^b = \frac{2.88 \times 10^{-1}}{7.20 \times 10^{-2}} = 4 $$
$$ 2^b = 4 \Rightarrow b = 2 $$

Thus, the overall order of the reaction is $$a + b = 1 + 2 = 3$$.

Answer: 3.

Equimolar mixture: $$x_1 = x_2 = 0.5$$.

By Raoult's law: $$P_{benzene} = 0.5 \times 80 = 40$$ Torr. $$P_{toluene} = 0.5 \times 24 = 12$$ Torr.

Total pressure = 52 Torr.

Mole fraction of toluene in vapour = $$12/52 = 3/13 = 0.2307... \approx 23 \times 10^{-2}$$.

The answer is $$\boxed{23}$$.

We need to find the ratio of the time for 99.9% completion to the time for 90% completion of a first-order reaction.

For a first-order reaction, the time for a given fraction of completion is given by $$t = \frac{2.303}{k} \log_{10}\left(\frac{C_0}{C}\right) = \frac{2.303}{k} \log_{10}\left(\frac{1}{1-x}\right)$$ where $$x$$ is the fraction of reaction completed and $$k$$ is the rate constant.

In the case of 99.9% completion, $$x = 0.999$$ so that $$1 - x = 0.001$$, hence $$t_{99.9\%} = \frac{2.303}{k} \log_{10}\left(\frac{1}{0.001}\right) = \frac{2.303}{k} \log_{10}(1000) = \frac{2.303}{k} \times 3$$

For 90% completion, $$x = 0.9$$ so that $$1 - x = 0.1$$, giving $$t_{90\%} = \frac{2.303}{k} \log_{10}\left(\frac{1}{0.1}\right) = \frac{2.303}{k} \log_{10}(10) = \frac{2.303}{k} \times 1$$

Thus, the ratio of these times is $$\frac{t_{99.9\%}}{t_{90\%}} = \frac{\frac{2.303 \times 3}{k}}{\frac{2.303 \times 1}{k}} = \frac{3}{1} = 3$$ and the answer is 3.

If concentration of A and B are doubled then x= (2^2)[A]*(2)[B] = 8[A][B], x=8

Since the curve between concentration and time is linear it is a zero order reaction. Hence, y =0

x+y = 8

The problem is to find the molality of a $$0.8$$ M $$H_2SO_4$$ solution whose density is $$1.06$$ g/cm$$^3$$; for $$H_2SO_4$$, the molar mass is $$2(1) + 32 + 4(16) = 98$$ g/mol.

Considering 1 litre (1000 mL) of this solution, the total mass is density × volume = $$1.06 \times 1000 = 1060$$ g.

With 0.8 mol of $$H_2SO_4$$ per litre, the mass of the solute is $$0.8 \times 98 = 78.4$$ g.

The mass of the solvent is then $$1060 - 78.4 = 981.6$$ g, or $$0.9816$$ kg.

The molality is given by $$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$$, so $$m = \frac{0.8}{0.9816} = 0.8150 \text{ mol/kg}$$.

Expressing in the required form: $$m = 815 \times 10^{-3}$$ m, which gives a numerical value of 815.

We need to find the osmotic pressure of a dilute solution at 283 K, given its osmotic pressure at 273 K.

Recall the relationship between osmotic pressure and temperature.

For a dilute solution, osmotic pressure is given by the van't Hoff equation:

$$\pi = CRT = \frac{n}{V}RT$$

where $$C$$ is the molar concentration, $$R$$ is the gas constant, and $$T$$ is the absolute temperature.

For the same solution (same $$C$$): $$\pi \propto T$$

Set up the proportion.

$$\frac{\pi_2}{\pi_1} = \frac{T_2}{T_1}$$

$$\pi_2 = \pi_1 \times \frac{T_2}{T_1} = 7 \times 10^5 \times \frac{283}{273}$$

Calculate.

$$\pi_2 = 7 \times 10^5 \times \frac{283}{273} = \frac{7 \times 283}{273} \times 10^5$$

$$= \frac{1981}{273} \times 10^5 = 7.2564 \times 10^5 \text{ Pa}$$

$$= 72.564 \times 10^4 \text{ Nm}^{-2} \approx 73 \times 10^4 \text{ Nm}^{-2}$$

The answer is 73.

Mass of X = 2 g. Mass of water = 1 mole = 18 g.

Total mass of solution = 2 + 18 = 20 g.

Mass percent of X = $$\frac{2}{20} \times 100 = 10\%$$

We need to determine the enthalpy changes for adsorption and micelle formation.

Enthalpy of adsorption ($$\Delta H_{ads}$$).

Adsorption is generally an exothermic process because when molecules adhere to a surface, bonds are formed, releasing energy. Both physical and chemical adsorption are exothermic.

Therefore, $$\Delta H_{ads} < 0$$.

Enthalpy of micelle formation ($$\Delta H_{mic}$$).

Micelle formation is primarily an entropy-driven process. When surfactant molecules form micelles, the ordered water molecules around the hydrophobic tails are released, increasing entropy significantly. The enthalpy change for micelle formation is typically slightly positive (endothermic) or near zero, but the large positive $$T\Delta S$$ term makes $$\Delta G$$ negative.

Therefore, $$\Delta H_{mic} > 0$$.

Match with options.

$$\Delta H_{ads} < 0$$ and $$\Delta H_{mic} > 0$$ corresponds to Option C.

The correct answer is Option C: $$\Delta H_{ads} < 0$$ and $$\Delta H_{mic} > 0$$.

The four statements given in the question deal with the fundamental properties of colloidal systems. We examine each one.

Statement (I) Lyophobic (solvent-hating) colloids cannot be prepared by simply mixing the dispersed phase with the dispersion medium.
Lyophobic particles have little affinity for the medium; therefore, special techniques such as electrical dispersion, ultrasonic dispersion or peptisation are required to break the bulk substance into colloidal-sized particles. Hence Statement (I) is true.

Statement (II) In an emulsion the dispersed phase as well as the dispersion medium are liquids.
Typical examples are milk (liquid fat dispersed in water) and cold cream (water dispersed in oil). By definition this is correct, so Statement (II) is true.

Statement (III) Micelles form whenever a surfactant is dissolved in any solvent at any temperature.
Micelle formation actually requires
  • a polar solvent, usually water,
  • surfactant concentration ≥ the critical micelle concentration (CMC), and
  • temperature ≥ the Kraft temperature.
If these conditions are not met, surfactant molecules remain as individual ions or molecules. Therefore Statement (III) is false.

Statement (IV) Tyndall effect is observed even when the dispersed phase and the dispersion medium have identical refractive indices.
The Tyndall effect (scattering of light by colloidal particles) becomes prominent only when there is a significant difference between the refractive indices of the two phases. If the indices are identical, scattering is negligible and the beam is invisible. Hence Statement (IV) is false.

Only Statements (I) and (II) are correct. Therefore, the correct choice is:

Option A which is: (I) and (II)

A surfactant molecule consists of two parts:

• A polar (hydrophilic) head
• A non-polar (hydrophobic) tail

The arrangement of surfactant molecules depends on the nature of the solvent.

In a polar solvent such as water, the polar heads interact with the solvent, while the non-polar tails remain away from it. Hence, the tails aggregate towards the center, forming normal micelles.

In a non-polar solvent, the arrangement is reversed. The non-polar tails interact readily with the non-polar solvent and point outward, whereas the polar heads avoid contact with the solvent and cluster together at the center.

Such an arrangement is called a reverse micelle.

Since the question involves a non-polar solvent, the surfactant molecules will form reverse micelles with:

• Polar heads directed inward
• Non-polar tails directed outward

Among the given diagrams, only diagram (a) represents this arrangement.

Hence, the correct answer is

$$\boxed{\text{Diagram (a)}}$$

Activated charcoal adsorbs a gas more strongly when the attractive van der Waals forces between its molecules are larger. For the noble gases those forces grow with increasing ease of liquefaction, or equivalently with higher critical temperature $$T_c$$. The critical temperatures are:

He : 5.2 K < Ne : 44.4 K < Ar : 150.9 K < Kr : 209.4 K.

Since krypton has the highest $$T_c$$, 1 g of activated charcoal will adsorb the greatest quantity of Kr. Hence Assertion A is true.

The Reason states that krypton has the highest critical volume $$V_c$$ and critical pressure $$P_c$$, but the lowest compressibility factor at the critical point $$Z_c$$. Experimental critical constants are:

He : $$P_c=2.3\ \text{atm},\;V_c=57\ \text{cm}^3\ \text{mol}^{-1},\;T_c=5.2\ \text{K}$$
Ne : $$P_c=26.3\ \text{atm},\;V_c=41\ \text{cm}^3\ \text{mol}^{-1},\;T_c=44.4\ \text{K}$$
Ar : $$P_c=48.3\ \text{atm},\;V_c=74.6\ \text{cm}^3\ \text{mol}^{-1},\;T_c=150.9\ \text{K}$$
Kr : $$P_c=55.0\ \text{atm},\;V_c=92.0\ \text{cm}^3\ \text{mol}^{-1},\;T_c=209.4\ \text{K}$$

Thus Kr indeed has the highest $$P_c$$ and $$V_c$$ among the four gases. However, the compressibility factor at the critical point is

$$Z_c=\frac{P_cV_c}{RT_c}$$

Using the data above gives $$Z_c\,$$: He 0.302, Ne 0.296, Ar 0.291, Kr 0.294. Krypton does not possess the lowest $$Z_c$$; argon’s value is slightly lower. Therefore the statement “$$Z_c$$ is lowest for krypton” is false, rendering Reason R false.

A true Assertion with a false Reason matches Option A.

Final answer : Option A.

The depression of freezing point is a colligative property that appears when a non-volatile solute is dissolved in a volatile solvent.

Case 1 : Statement A
Statement A says: “Vapour pressure of the solution is less than that of pure solvent.”
According to Raoult’s law, addition of a non-volatile solute lowers the vapour pressure of the solvent:
$$P_\text{solution}=x_\text{solvent}\,P^\circ_\text{solvent}$$
Since $$x_\text{solvent}\lt1$$, $$P_\text{solution}\lt P^\circ_\text{solvent}$$. Hence Statement A is correct.

Case 2 : Statement B
Statement B is the opposite of Statement A, therefore it is false.

Case 3 : Statement C
Statement C says: “Only solute molecules solidify at the freezing point.”
At the freezing point an equilibrium exists between solid solvent and liquid solution. The solid phase that separates out is pure solvent; solute molecules remain in the liquid phase. Hence Statement C is false.

Case 4 : Statement D
Statement D says: “Only solvent molecules solidify at the freezing point.”
As explained above, the solid that forms is pure solvent. Hence Statement D is correct.

Thus, the correct statements are A and D only.

Option A (A and D only) is the correct answer.

In the depression of freezing point experiment, we analyze each statement:

Statement A: Vapour pressure of the solution is less than that of pure solvent.

This is TRUE. According to Raoult's law, the addition of a non-volatile solute lowers the vapour pressure of the solvent.

Statement B: Vapour pressure of the solution is more than that of pure solvent.

This is FALSE. It contradicts Raoult's law.

Statement C: Only solute molecules solidify at the freezing point.

This is FALSE. At the freezing point of a solution, it is the solvent that begins to solidify (separate out as pure solid), not the solute.

Statement D: Only solvent molecules solidify at the freezing point.

This is TRUE. When a solution reaches its freezing point, only the pure solvent crystallises out.

The correct statements are A and D.

The correct answer is Option A: A and D only.

We need to match List I with List II for concepts related to solutions.

A. van't Hoff factor, $$i$$:

The van't Hoff factor is defined as:

$$i = \frac{\text{Normal molar mass}}{\text{Abnormal molar mass}}$$

This matches with III.

B. $$K_f$$:

$$K_f$$ is the cryoscopic constant (molal depression constant for freezing point).

This matches with I.

C. Solutions with same osmotic pressure:

Solutions having the same osmotic pressure are called isotonic solutions.

This matches with II.

D. Azeotropes:

Azeotropes are solutions that have the same composition in the liquid and vapour phase. They boil at a constant temperature and the vapour has the same composition as the liquid.

This matches with IV (solutions with same composition of vapour above it).

The correct answer is Option A: A-III, B-I, C-II, D-IV.

We need to match List-I with List-II regarding osmosis-related phenomena.

A. Osmosis: The spontaneous flow of solvent molecules through a semi-permeable membrane from pure solvent (or dilute solution) towards the solution side.

This matches with III: Solvent molecules pass through semi permeable membrane towards solution side.

B. Reverse osmosis: When external pressure greater than osmotic pressure is applied, solvent molecules move from solution side to solvent side through the membrane.

This matches with I: Solvent molecules pass through semi permeable membrane towards solvent side.

C. Electro osmosis: The movement of dispersion medium (solvent) under the influence of an applied electric field through a membrane.

This matches with IV: Dispersion medium moves in an electric field.

D. Electrophoresis: The movement of charged colloidal particles under the influence of applied electric potential towards oppositely charged electrodes.

This matches with II: Movement of charged colloidal particles under the influence of applied electric potential towards oppositely charged electrodes.

The correct matching is: A-III, B-I, C-IV, D-II

The correct answer is Option B: A-III, B-I, C-IV, D-II.

The recorded answer is Option D (code 4): A-I, B-III, C-II, D-IV. However, osmosis is movement towards solution side (not solvent side), so A should match III. Our answer (Option B) is correct. Saving for review.

We need to evaluate both statements about colloidal particles and determine their truth values. For colloidal particles, the values of colligative properties are of small order as compared to values shown by true solutions at the same concentration. This is TRUE. Colligative properties (such as osmotic pressure, boiling point elevation, freezing point depression, and relative lowering of vapour pressure) depend on the number of solute particles in solution, not on their nature or size. The key formula for osmotic pressure, for example, is:

where $$n$$ is the number of moles of solute particles. In a colloidal solution, each colloidal particle is actually an aggregate of many individual molecules or ions. Therefore, at the same mass concentration, a colloidal solution has far fewer particles than a true solution. Since fewer particles means smaller colligative effects, colloidal solutions show much smaller values of colligative properties compared to true solutions at the same concentration.

For colloidal particles, the potential difference between the fixed layer and the diffused layer of same charges is called the electrokinetic potential or zeta potential. This is TRUE. In a colloidal system, each colloidal particle carries an electric charge. This charge attracts counter-ions from the surrounding medium, forming what is known as the electrical double layer:
Fixed layer (Stern layer): A layer of ions firmly attached to the surface of the colloidal particle.
Diffused layer: A layer of ions that are loosely associated and extend into the bulk of the solution.
The potential difference between the fixed (Stern) layer and the diffused layer is called the zeta potential (or electrokinetic potential). This zeta potential is crucial because it determines the stability of the colloidal system. A high zeta potential (positive or negative) indicates a stable colloid, while a low zeta potential means the particles can come close enough to aggregate and coagulate.

Both Statement I and Statement II are true. The correct answer is Option 3.

We analyze each statement:

Statement I: For colloidal particles, the values of colligative properties are of small order as compared to values shown by true solutions at the same concentration.

This is TRUE. Colloidal particles are aggregates of many molecules, so the effective number of solute particles is much less than in a true solution of the same concentration. Since colligative properties depend on the number of particles, they are smaller for colloids.

Statement II: For colloidal particles, the potential difference between the fixed layer and the diffused layer of same charges is called the electrokinetic potential or zeta potential.

This is TRUE. The zeta potential is defined as the potential difference between the fixed (Stern) layer and the diffused layer of the electrical double layer surrounding a colloidal particle.

Both Statement I and Statement II are true.

The correct answer is Option C: Both Statement I and Statement II are true.

We need to find the weight of glucose to dissolve in 100 g of water to lower vapour pressure by 0.20 mm Hg.

$$P^\circ = 54.2$$ mm Hg, $$\Delta P = 0.20$$ mm Hg, molar mass of glucose = 180 g/mol, molar mass of water = 18 g/mol.

Apply Raoult's law for dilute solutions.

For a dilute solution, the relative lowering of vapour pressure is:

$$ \frac{\Delta P}{P^\circ} = \frac{n_{\text{solute}}}{n_{\text{solvent}}} $$

(For dilute solutions, $$n_{\text{solute}} + n_{\text{solvent}} \approx n_{\text{solvent}}$$, so this simplification is valid.)

Express moles in terms of given quantities.

$$n_{\text{solute}} = \frac{w}{180}$$ (where $$w$$ is the weight of glucose in grams)

$$n_{\text{solvent}} = \frac{100}{18}$$

Substitute and solve.

$$ \frac{0.20}{54.2} = \frac{w/180}{100/18} = \frac{w}{180} \times \frac{18}{100} = \frac{w}{1000} $$

$$ w = \frac{0.20 \times 1000}{54.2} = \frac{200}{54.2} = 3.69\;\text{g} $$

The weight of glucose required is 3.69 g.

The correct answer is Option 2: 3.69 g.

The Freundlich adsorption isotherm is: $$\frac{x}{m} = kP^{1/n}$$

This can be represented graphically in several ways:

A. $$x/m$$ vs $$P$$: This gives a curve (not linear) which is characteristic of Freundlich adsorption. ✓

B. Taking log on both sides: $$\log\frac{x}{m} = \log k + \frac{1}{n}\log P$$. Plot of $$\log(x/m)$$ vs $$\log P$$ is a straight line with slope $$1/n$$ and intercept $$\log k$$. ✓

C. $$x/m$$ vs $$c$$ (concentration): This doesn't directly represent the Freundlich isotherm equation (which uses pressure for gases). ✗

D. $$x/m$$ vs $$P^{1/n}$$: From the equation, $$x/m = k \cdot P^{1/n}$$, this is a straight line passing through the origin with slope $$k$$. ✓

The correct representations are A, B, D only.

When a lyophilic sol (solvent-loving colloid) is added to a lyophobic sol (solvent-hating colloid), the lyophilic sol particles form a protective layer (film) around the lyophobic sol particles.

This process is called protective action of lyophilic colloids. The lyophilic colloid gets adsorbed on the surface of lyophobic colloid particles, forming a protective film. This prevents the coagulation of the lyophobic sol and makes it more stable.

The answer is Option 1: Film of lyophilic sol is formed over lyophobic sol.

For the first case, let the mass of ethylene glycol be (w_1) g. Since the solution is (0.25) molal,

$$0.25=\frac{w_1/62}{(500-w_1)/1000}.$$

On solving,

$$0.25=\frac{1000w_1}{62(500-w_1)},$$

$$15.5(500-w_1)=1000w_1,$$

$$7750-15.5w_1=1000w_1,$$

$$7750=1015.5w_1,$$

$$w_1\approx7.63\text{ g}.$$

For the second case, the solution is (0.25) M and its volume is (250) mL (=0.250) L.

The number of moles of ethylene glycol is

$$0.25\times0.250=0.0625\text{ mol}.$$

Hence, the required mass is

$$w_2=0.0625\times62=3.875\text{ g}.$$

Therefore, the required mass ratio is

$$\frac{w_1}{w_2}=\frac{7.63}{3.875}\approx1.97\approx2.$$

Hence, the mass ratio is

$$\boxed{2:1}.$$

Assertion A: The diameter of colloidal particles should not be much smaller than the wavelength of light to show the Tyndall effect.

This is true. For the Tyndall effect (scattering of light), the particle size must be comparable to the wavelength of light. If particles are much smaller than the wavelength, they won't scatter light effectively.

Reason R: Light scatters in all directions when the size of particles is large enough.

This is true. When particles are of appropriate size, they scatter light in all directions, making the beam visible (Tyndall effect).

R correctly explains why the particle size matters for the Tyndall effect - the scattering occurs when particles are large enough relative to the wavelength.

Both A and R are correct and R is the correct explanation of A.

This matches option 4.

Matching the items:

A. Physisorption: Characterized by low enthalpy of adsorption, typically in the range of 20-40 kJ mol$$^{-1}$$. → II

B. Chemisorption: Forms a single layer (monolayer) on the surface because chemical bonds are formed between the adsorbate and specific surface sites. → I

C. N$$_2$$(g) + 3H$$_2$$(g) $$\xrightarrow{Fe(s)}$$ 2NH$$_3$$(g): This is the Haber process where iron acts as a solid catalyst with gaseous reactants — a classic example of heterogeneous catalysis. → IV

D. Analytical application of adsorption: Chromatography is the most important analytical application based on differential adsorption. → III

The correct matching is: A-II, B-I, C-IV, D-III.

When FeCl₃ is added to NaOH solution, Fe(OH)₃ colloidal solution is formed. Since NaOH is in excess, the colloid adsorbs OH⁻ ions and becomes negatively charged.

By the Hardy-Schulze rule, for coagulation of a negatively charged colloid, we need cations with the highest charge. Higher the charge of the coagulating ion, greater the coagulating power.

Examining the options:

Option A: AlCl₃ → provides Al³⁺ (charge +3) — highest coagulating power

Option B: Na₂SO₄ → provides Na⁺ (charge +1) — lowest coagulating power

Option C: Ca₃(PO₄)₂ → provides Ca²⁺ (charge +2)

Option D: CaCl₂ → provides Ca²⁺ (charge +2)

Al³⁺ has the highest charge (+3), so it will coagulate the negatively charged colloid at the fastest rate, even considering the concentration differences.

The answer is Option A: 10 mL of 0.2 mol dm⁻³ AlCl₃.

Blood is a negatively charged colloidal solution. Fe³⁺ ions (from FeCl₃), being trivalent cations, are very effective at coagulating negatively charged colloids (Hardy-Schulze rule).

The correct answer is Option 3.

We are given four gases A, B, C, and D with critical temperatures 5.3 K, 33.2 K, 126.0 K, and 154.3 K respectively, and we need to determine the correct order of their adsorption on a fixed amount of charcoal. The extent of adsorption of a gas on a solid surface increases with the ease of liquefaction of the gas, so a gas with a higher critical temperature is more easily liquefied and therefore more readily adsorbed.

The critical temperatures of the four gases are:
A = 5.3 K
B = 33.2 K
C = 126.0 K
D = 154.3 K

Arranging these in decreasing order of critical temperature (and hence decreasing order of adsorption) gives
$$D \, (154.3 \, K) \; > \; C \, (126.0 \, K) \; > \; B \, (33.2 \, K) \; > \; A \, (5.3 \, K)$$

Therefore, the correct order of adsorption is D > C > B > A.

The correct answer is Option C.

We have 200 g of a 25% sugar solution and 500 g of a 40% sugar solution. The mass of sugar in the first solution is $$\frac{25}{100} \times 200 = 50$$ g, and in the second solution is $$\frac{40}{100} \times 500 = 200$$ g.

The total sugar is $$50 + 200 = 250$$ g and the total mass of solution is $$200 + 500 = 700$$ g.

Now, the mass percentage is:

$$\text{Mass \%} = \frac{250}{700} \times 100 = 35.71\%$$

Rounding to the nearest integer gives approximately $$36\%$$.

So, the answer is $$36$$.

We have a 3M solution of NaCl with density = 1.0 g/mL. The molar mass of NaCl = 23 + 35.5 = 58.5 g/mol.

In 1 litre of solution, the mass of solution = 1000 mL $$\times$$ 1.0 g/mL = 1000 g, and the moles of NaCl = 3 mol, giving mass of NaCl = 3 $$\times$$ 58.5 = 175.5 g.

Now, the mass of solvent (water) is:

$$m_{water} = 1000 - 175.5 = 824.5 \text{ g} = 0.8245 \text{ kg}$$

Molality is defined as moles of solute per kg of solvent:

$$\text{Molality} = \frac{3}{0.8245} = 3.6386 \text{ m}$$

In terms of $$\times 10^{-2}$$ m:

$$\text{Molality} = 363.86 \times 10^{-2} \text{ m} \approx 364 \times 10^{-2} \text{ m}$$

So, the answer is $$364$$.

Explanation

According to the kinetic theory of gases,

The root mean square speed is

$$v_{rms}=\sqrt{\frac{3RT}{M}}$$

The most probable speed is

$$v_{mp}=\sqrt{\frac{2RT}{M}}$$

The problem states that

$$(v_{rms})_X=(v_{mp})_Y$$

Substituting the respective expressions,

$$\sqrt{\frac{3RT_X}{M_X}}=\sqrt{\frac{2RT_Y}{M_Y}}$$

Squaring both sides and cancelling $$R$$,

$$\frac{3T_X}{M_X}=\frac{2T_Y}{M_Y}$$

Substituting the given values,

$$\frac{3\times600}{40}=\frac{2\times90}{M_Y}$$

$$\frac{1800}{40}=\frac{180}{M_Y}$$

$$45=\frac{180}{M_Y}$$

$$M_Y=\frac{180}{45}$$

$$M_Y=4\text{ g mol}^{-1}$$

Hence, the molar mass of gas $$Y$$ is

$$4\text{ g mol}^{-1}$$

Therefore, the correct answer is **4**.

The boiling point elevation constant (ebullioscopic constant) is given by:

$$K_b = \frac{R T_b^2 M}{1000 \, \Delta H_{\text{vap}}}$$

where $$R$$ is the gas constant, $$T_b$$ is the boiling point (in Kelvin), $$M$$ is the molar mass of the solvent, and $$\Delta H_{\text{vap}}$$ is the enthalpy of vaporization.

Since both solvents X and Y have the same molecular weight, $$M$$ and $$R$$ cancel in the ratio:

$$\frac{K_{b,X}}{K_{b,Y}} = \frac{T_{b,X}^2}{T_{b,Y}^2} \times \frac{\Delta H_{\text{vap},Y}}{\Delta H_{\text{vap},X}}$$

We have $$\frac{T_{b,X}}{T_{b,Y}} = \frac{2}{1}$$ and $$\frac{\Delta H_{\text{vap},X}}{\Delta H_{\text{vap},Y}} = \frac{1}{2}$$. Substituting:

$$\frac{K_{b,X}}{K_{b,Y}} = \left(\frac{2}{1}\right)^2 \times \frac{2}{1} = 4 \times 2 = 8$$

So $$K_{b,X} = 8 \times K_{b,Y}$$, meaning $$m = 8$$.

So, the answer is $$8$$.

We need to find the molality of a $$10\%$$ (v/V) solution of dibromine (Br$$_2$$) in CCl$$_4$$.

Interpret 10% (v/V).

$$10\%$$ (v/V) means 10 mL of Br$$_2$$ is present in 100 mL of solution.

Find the mass of Br$$_2$$ (solute).

Mass of Br$$_2$$ = Volume $$\times$$ Density = $$10 \times 3.2 = 32$$ g

Moles of Br$$_2$$ = $$\frac{32}{160} = 0.2$$ mol

Find the mass of CCl$$_4$$ (solvent).

Volume of CCl$$_4$$ = Total volume $$-$$ Volume of Br$$_2$$ = $$100 - 10 = 90$$ mL

Mass of CCl$$_4$$ = $$90 \times 1.6 = 144$$ g = $$0.144$$ kg

Calculate molality.

$$ \text{Molality} = \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}} = \frac{0.2}{0.144} = 1.3889 \text{ mol/kg} $$

Express in the required form.

$$x \times 10^{-2} = 1.3889$$

$$x = 138.89 \approx 139$$ (nearest integer)

The answer is $$\boxed{139}$$.

We need to find the volume of dissolved CO$$_2$$ at STP.

The solution has a volume of 300 mL = 0.3 L and a concentration of 0.2 M, so the moles of CO$$_2$$ are 0.3 × 0.2 = 0.06 mol.

At STP, molar volume = 22.7 L/mol, so the volume is 0.06 × 22.7 = 1.362 L = 1362 mL.

The volume of dissolved CO$$_2$$ at STP is $$\boxed{1362}$$ mL.

We are given the coagulating values of two electrolytes for $$As_2S_3$$ sol:

Coagulating value of $$AlCl_3$$ = 0.09

Coagulating value of $$NaCl$$ = 50.04

Coagulating power is the reciprocal of coagulating value — a lower coagulating value means a higher coagulating power (less electrolyte is needed to cause coagulation). So we have:

$$\text{Coagulating power} = \frac{1}{\text{Coagulating value}}$$

Now, coagulating power of $$AlCl_3$$:

$$P_{\text{AlCl}_3} = \frac{1}{0.09}$$

And coagulating power of $$NaCl$$:

$$P_{\text{NaCl}} = \frac{1}{50.04}$$

Since the coagulating power of $$AlCl_3$$ is $$x$$ times the coagulating power of $$NaCl$$:

$$x = \frac{P_{\text{AlCl}_3}}{P_{\text{NaCl}}} = \frac{1/0.09}{1/50.04} = \frac{50.04}{0.09} = 556$$

This is consistent with the Hardy-Schulze rule: $$Al^{3+}$$ (trivalent cation) has much greater coagulating power than $$Na^+$$ (monovalent cation) for the negatively charged $$As_2S_3$$ sol.

Hence, the value of $$x$$ is $$556$$.

To compare the osmotic pressures of two solutions at the same temperature, we use the relation

$$\pi=iMRT.$$

Since (R) and (T) are constant, two solutions will have the same osmotic pressure if they have the same value of (iM), where (i) is the van't Hoff factor and (M) is the molarity.

For Pair A,

$$C_2H_5OH:\ i=1,\quad iM=1\times0.500=0.50,$$

$$KBr:\ i=2,\quad iM=2\times0.25=0.50.$$

Hence, both solutions have the same osmotic pressure.

For Pair B,

$$K_4[Fe(CN)_6]:\ i=5,\quad iM=5\times0.100=0.50,$$

$$FeSO_4\cdot(NH_4)_2SO_4:\ i=5,\quad iM=5\times0.100=0.50.$$

Hence, both solutions have the same osmotic pressure.

For Pair C,

$$K_4[Fe(CN)_6]:\ iM=5\times0.05=0.25,$$

$$NaCl:\ iM=2\times0.25=0.50.$$

Since the values are different, these solutions do not have the same osmotic pressure.

For Pair D,

$$NaCl:\ iM=2\times0.15=0.30,$$

$$BaCl_2:\ iM=3\times0.10=0.30.$$

Hence, both solutions have the same osmotic pressure.

For Pair E,

$$KCl\cdot MgCl_2\cdot6H_2O:\ i=5,\quad iM=5\times0.02=0.10,$$

$$KCl:\ i=2,\quad iM=2\times0.05=0.10.$$

Hence, both solutions have the same osmotic pressure.

Therefore, Pairs A, B, D, and E have equal osmotic pressures, while Pair C does not. Hence, the number of pairs having the same osmotic pressure is

$$\boxed{4}.$$

30% (w/v) glucose solution means 30 g glucose in 100 mL solution.

Density = 1.2 g/cm³, so mass of 100 mL solution = 120 g.

Mass of water (solvent) = 120 - 30 = 90 g.

Moles of glucose = $$\frac{30}{180} = \frac{1}{6}$$ mol

Moles of water = $$\frac{90}{18} = 5$$ mol

Using Raoult's law:

$$\frac{P^0 - P}{P^0} = x_{solute} = \frac{1/6}{1/6 + 5} = \frac{1/6}{31/6} = \frac{1}{31}$$

$$P = P^0\left(1 - \frac{1}{31}\right) = 24 \times \frac{30}{31} = \frac{720}{31} \approx 23.23$$

Nearest integer: $$P \approx 23$$ mm Hg

This matches the answer key value of $$\mathbf{23}$$.

We need to find the depression in freezing point when 5 g of acetic acid (20% dissociated) is added to 500 mL of water.

Calculate the moles of acetic acid.

Molar mass of CH$$_3$$COOH = 12 + 3(1) + 12 + 16 + 16 + 1 = 60 g/mol.

$$n = \frac{5}{60} = \frac{1}{12} \text{ mol}$$

Calculate the van't Hoff factor.

Acetic acid dissociates as: CH$$_3$$COOH $$\rightleftharpoons$$ CH$$_3$$COO$$^-$$ + H$$^+$$

With degree of dissociation $$\alpha = 0.20$$:

$$i = 1 + \alpha = 1 + 0.20 = 1.2$$

Calculate the molality.

Mass of water = 500 mL $$\times$$ 1 g/cm$$^3$$ = 500 g = 0.5 kg.

$$m = \frac{1/12}{0.5} = \frac{1}{6} \text{ mol/kg}$$

Calculate the depression in freezing point.

$$\Delta T_f = i \cdot K_f \cdot m = 1.2 \times 1.86 \times \frac{1}{6}$$

$$\Delta T_f = 1.2 \times 0.31 = 0.372 \text{ °C}$$

Express in the required form.

$$\Delta T_f = 0.372 \text{ °C} = 372 \times 10^{-3} \text{ °C}$$

The correct answer is 372.

To determine the molecular mass of a non-volatile solute from the elevation in boiling point, note that the solute mass is 2 g and the solvent mass is 20 g = 0.020 kg. The boiling point increase is $$\Delta T_b = 373.52 - 373 = 0.52$$ K, and the ebullioscopic constant is $$K_b = 0.52$$ K kg mol$$^{-1}$$.

Using the relation $$\Delta T_b = K_b \times m = K_b \times \frac{w_2}{M_2 \times w_1}$$ and substituting the values gives:

$$0.52 = 0.52 \times \frac{2}{M_2 \times 0.020}$$

Rearranging yields $$1 = \frac{2}{0.020 \times M_2} = \frac{100}{M_2}$$, so $$M_2 = 100$$ g/mol.

Therefore, the molecular mass of the solute is $$\boxed{100}$$ g mol$$^{-1}$$.

For an ideal binary liquid mixture, Raoult's law is

$$P_{\text{Total}} = x_A P_A^\circ + x_B P_B^\circ$$

From the first set of data,

$$0.7P_A^\circ + 0.3P_B^\circ = 350 \qquad ...(1)$$

From the second set of data,

$$0.2P_A^\circ + 0.8P_B^\circ = 410 \qquad ...(2)$$

Multiplying equation (1) by 8,

$$5.6P_A^\circ + 2.4P_B^\circ = 2800$$

Multiplying equation (2) by 3,

$$0.6P_A^\circ + 2.4P_B^\circ = 1230$$

Subtracting,

$$\left(5.6-0.6\right)P_A^\circ + \left(2.4-2.4\right)P_B^\circ = 2800-1230$$

$$5P_A^\circ = 1570$$

$$P_A^\circ = \frac{1570}{5}$$

$$P_A^\circ = 314\ \text{mm Hg}$$

Hence, the vapour pressure of pure liquid A is

$$\boxed{314\ \text{mm Hg}}$$

Finding the moles of each gas:

$$ n_X = \frac{0.6}{20} = 0.03 \text{ mol} $$

$$ n_Y = \frac{0.45}{45} = 0.01 \text{ mol} $$

Total moles: $$n_{\text{total}} = 0.03 + 0.01 = 0.04$$ mol

Mole fraction of gas X:

$$ \chi_X = \frac{n_X}{n_{\text{total}}} = \frac{0.03}{0.04} = 0.75 $$

Partial pressure of gas X (by Dalton's law):

$$ P_X = \chi_X \times P_{\text{total}} = 0.75 \times 740 = 555 \text{ mm of Hg} $$

For isotonic solutions, the osmotic pressures are equal:

$$i_1 C_1 = i_2 C_2$$

For glucose (non-electrolyte): $$i_2 = 1$$, $$C_2 = 0.01$$ M

For K$$_2$$SO$$_4$$: $$C_1 = 0.004$$ M

$$i_1 \times 0.004 = 1 \times 0.01$$

$$i_1 = \frac{0.01}{0.004} = 2.5$$

K$$_2$$SO$$_4$$ dissociates as: $$\text{K}_2\text{SO}_4 \rightarrow 2\text{K}^+ + \text{SO}_4^{2-}$$

Number of ions $$n = 3$$

Using the van't Hoff factor formula:

$$i = 1 + \alpha(n - 1)$$

$$2.5 = 1 + \alpha(3 - 1)$$

$$2.5 = 1 + 2\alpha$$

$$\alpha = \frac{1.5}{2} = 0.75$$

Percentage dissociation = $$0.75 \times 100 = 75\%$$

We have 25 mL of KCl solution titrated with 20 mL of 1M $$AgNO_3$$ solution. Given: $$K_f = 2.0$$ K kg $$mol^{-1}$$, 100% ionization, density of solution = 1 g/mL.

The reaction is: $$KCl + AgNO_3 \rightarrow AgCl\downarrow + KNO_3$$

Moles of $$AgNO_3$$ used = 0.020 $$\times$$ 1 = 0.02 mol. Since the ratio is 1:1, moles of KCl = 0.02 mol.

Now, mass of solution = 25 $$\times$$ 1 = 25 g, and mass of KCl = 0.02 $$\times$$ 74.5 = 1.49 g. So the mass of solvent (water) = 25 - 1.49 = 23.51 g = 0.02351 kg.

The molality is:

$$m = \frac{0.02}{0.02351} = 0.8507 \text{ mol/kg}$$

For KCl with 100% ionization ($$KCl \rightarrow K^+ + Cl^-$$), the van't Hoff factor $$i = 2$$.

So the depression in freezing point is:

$$\Delta T_f = i \times K_f \times m = 2 \times 2.0 \times 0.8507 = 3.40 \text{ K}$$

Rounding to the nearest integer, the answer is $$3$$ K.

Stock: 5g NaOH (MW=40) = 0.125 mol in 450 mL. M = 0.125/0.45 = 0.2778 M.

For 500 mL of 0.1 M: moles = 0.05. V = 0.05/0.2778 = 0.180 L = 180 mL.

For $$MgCl_2$$,

$$MgCl_2 \rightarrow Mg^{2+}+2Cl^-$$

The total number of ions produced is $3$. Therefore, the van't Hoff factor is

$$i=1+\alpha(n-1)$$

$$i=1+0.8(3-1)=2.6$$

A $$1.0$$ molal solution contains $$1$$ mol of $$MgCl_2$$ dissolved in $$1000$$ g of water.

The moles of water are

$$n_{\text{water}}=\frac{1000}{18}=55.56\ \text{mol}$$

Using Raoult's law,

$$\frac{P^\circ-P}{P^\circ}=\frac{i,n_2}{n_1+i,n_2}$$

Substituting the given values,

$$\frac{50-P}{50}=\frac{2.6\times1}{55.56+2.6}=\frac{2.6}{58.16}\approx0.0447$$

Hence,

$$50-P=50\times0.0447\approx2.24\ \text{mm Hg}$$

$$P=50-2.24\approx47.76\ \text{mm Hg}$$

Therefore,

$$P\approx48\ \text{mm Hg}$$

The calculated value of approximately $$2\ \text{mm Hg}$$ represents the lowering of vapor pressure $$(\Delta P)$$, whereas the vapor pressure of the solution is approximately $$48\ \text{mm Hg}$$.

Using the osmotic pressure formula:

$$ \pi = \frac{n}{V}RT = \frac{w}{MV}RT $$

Rearranging for molar mass $$M$$:

$$ M = \frac{wRT}{\pi V} $$

Given: $$w = 2.5$$ g, $$V = 250.0$$ mL $$= 0.25$$ L, $$T = 27°C = 300$$ K, $$R = 0.083$$ L bar K$$^{-1}$$ mol$$^{-1}$$

Converting osmotic pressure to bar: $$\pi = 400$$ Pa $$= \frac{400}{10^5}$$ bar $$= 0.004$$ bar

$$ M = \frac{2.5 \times 0.083 \times 300}{0.004 \times 0.25} = \frac{62.25}{0.001} = 62250 \text{ g mol}^{-1} $$

Two solutions are isotonic (have the same osmotic pressure) at the same temperature when they have the same effective concentration, i.e., $$i \times M$$ (van't Hoff factor times molarity) is equal.

For Pair A, 1 M $$\text{NaCl} \to \text{Na}^+ + \text{Cl}^-$$ with $$i = 2$$ so $$i \times M = 2 \times 1 = 2$$; 2 M urea is a non-electrolyte with $$i = 1$$ so $$i \times M = 1 \times 2 = 2$$. Since $$2 = 2$$, they are isotonic.

For Pair B, 1 M $$\text{CaCl}_2 \to \text{Ca}^{2+} + 2\text{Cl}^-$$ with $$i = 3$$ so $$i \times M = 3 \times 1 = 3$$; 1.5 M $$\text{KCl} \to \text{K}^+ + \text{Cl}^-$$ with $$i = 2$$ so $$i \times M = 2 \times 1.5 = 3$$. Since $$3 = 3$$, they are isotonic.

For Pair C, 1.5 M $$\text{AlCl}_3 \to \text{Al}^{3+} + 3\text{Cl}^-$$ with $$i = 4$$ so $$i \times M = 4 \times 1.5 = 6$$; 2 M $$\text{Na}_2\text{SO}_4 \to 2\text{Na}^+ + \text{SO}_4^{2-}$$ with $$i = 3$$ so $$i \times M = 3 \times 2 = 6$$. Since $$6 = 6$$, they are isotonic.

For Pair D, 2.5 M $$\text{KCl}$$ has $$i = 2$$ so $$i \times M = 2 \times 2.5 = 5$$; 1 M $$\text{Al}_2(\text{SO}_4)_3 \to 2\text{Al}^{3+} + 3\text{SO}_4^{2-}$$ with $$i = 5$$ so $$i \times M = 5 \times 1 = 5$$. Since $$5 = 5$$, they are isotonic.

All four pairs are isotonic; therefore the number of isotonic pairs is 4.

The concentration in ppm by mass is given by

$$\text{ppm}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times10^6$$

The mass of chloroform is

$$\text{Mass of }CHCl_3=671.141\times1.49\approx1000\ \text{g}$$

Thus, the mass of the solvent is approximately $1000\ \text{g}$.

The molar mass of $CH_2Cl_2$ is

$$12+(2\times1)+(2\times35.5)=85\ \text{g mol}^{-1}$$

Since the solution is very dilute, its volume is approximately equal to the solvent volume.

$$V=671.141\ \text{mL}=0.671141\ \text{L}$$

The moles of $CH_2Cl_2$ are

$$n=MV=(2.6\times10^{-3})\times0.671141\approx1.745\times10^{-3}\ \text{mol}$$

Hence, the mass of $CH_2Cl_2$ is

$$m=n\times M=(1.745\times10^{-3})\times85\approx0.1483\ \text{g}$$

Substituting into the ppm expression,

$$\text{ppm}=\frac{0.1483}{1000}\times10^6=148.3$$

Therefore, the concentration of the solution is

$$\boxed{148\ \text{ppm}}$$

The concentration of a solution is the amount of solute present in a given amount of solvent or solution. Therefore, only those quantities that express a ratio of solute to solvent/solution can be used as concentration units.

Mass percent $$(w/w\%)$$ is defined as the mass of solute per 100 parts by mass of solution and is a valid concentration term.

Mole is the SI unit of amount of substance. It represents only the quantity of particles and does not express any ratio. Hence, it is not a concentration term.

Mole fraction $$(x)$$ is the ratio of the moles of a component to the total moles present in the solution and is a valid concentration term.

Molarity $$(M)$$ is the number of moles of solute present in one litre of solution and is a concentration term.

ppm (parts per million) expresses the number of parts of solute per million parts of solution and is commonly used for very dilute solutions. It is also a concentration term.

Molality $$(m)$$ is the number of moles of solute present per kilogram of solvent and is a concentration term.

Thus, the valid concentration terms are:

• Mass percent

• Mole fraction

• Molarity

• ppm

• Molality

"Mole" is the only quantity that does not represent concentration.

Hence, the number of terms used to express concentration is

$$\boxed{5}$$

Coagulating value is defined as the millimoles of electrolyte required to cause coagulation of 1 L of sol.

Millimoles of NaCl used = 20 × 0.5 = 10 mmol

Volume of sol = 200 mL = 0.2 L

Coagulating value = $$\frac{10}{0.2} = 50$$ mmol/L

The coagulating value is $$\mathbf{50}$$.

We need to find the molar mass of a protein given its osmotic pressure data.

Volume = 300 cm$$^3$$ = 0.3 L, mass of protein = 0.63 g, Temperature = 300 K, Osmotic pressure $$\pi = 1.29$$ mbar = $$1.29 \times 10^{-3}$$ bar, $$R = 0.083$$ L bar K$$^{-1}$$ mol$$^{-1}$$.

The van't Hoff equation for osmotic pressure is:

$$\pi = CRT = \frac{n}{V}RT = \frac{m}{MV}RT$$

where $$C$$ is molar concentration, $$n$$ is moles of solute, $$m$$ is mass of solute, $$M$$ is molar mass, $$V$$ is volume in liters.

$$M = \frac{mRT}{\pi V}$$

$$M = \frac{0.63 \times 0.083 \times 300}{1.29 \times 10^{-3} \times 0.3}$$

$$0.63 \times 0.083 = 0.05229$$

$$0.05229 \times 300 = 15.687$$

$$1.29 \times 10^{-3} \times 0.3 = 3.87 \times 10^{-4}$$

Divide.

$$M = \frac{15.687}{3.87 \times 10^{-4}} = \frac{15.687}{0.000387} \approx 40535 \text{ g/mol}$$

The molar mass of the protein is 40535 g mol$$^{-1}$$.

We are given 12 g of non-electrolyte A dissolved in 1000 mL of water, exerting the same osmotic pressure as 0.05 M glucose at the same temperature. The empirical formula of A is CH$$_2$$O.

For non-electrolytes at the same temperature: $$\pi = CRT$$.

Since both solutions have the same osmotic pressure:

Volume of solution $$\approx 1000 \text{ mL} = 1 \text{ L}$$ (dilute solution).

Empirical formula mass of CH$$_2$$O = $$12 + 2 + 16 = 30$$ g/mol.

Molecular formula = C$$_8$$H$$_{16}$$O$$_8$$.

The molecular mass of A is $$240$$ g.

For a weak monobasic acid with degree of dissociation $$\alpha = 0.3$$:

$$HA \rightleftharpoons H^+ + A^-$$

The van't Hoff factor: $$i = 1 + \alpha = 1 + 0.3 = 1.3$$

The observed freezing point depression: $$\Delta T_f = i \cdot K_f \cdot m$$

The expected (without dissociation): $$\Delta T_{f,expected} = K_f \cdot m$$

The observed depression is $$i = 1.3$$ times the expected depression.

The observed depression is $$(1.3 - 1)/1 \times 100 = 30\%$$ higher than expected.

The observed freezing point depression will be 30% higher than expected.

Given: 29.25% NaCl and 19% MgCl₂ by weight. K_b = 0.52 K kg/mol.

Consider 100 g of sea water: 29.25 g NaCl, 19 g MgCl₂, solvent = 100 - 29.25 - 19 = 51.75 g = 0.05175 kg.

Moles: NaCl = 29.25/58.5 = 0.5 mol, MgCl₂ = 19/95 = 0.2 mol.

Van't Hoff factors: NaCl → Na⁺ + Cl⁻ (i = 2), MgCl₂ → Mg²⁺ + 2Cl⁻ (i = 3).

Total molality of particles: $$\frac{0.5 \times 2 + 0.2 \times 3}{0.05175} = \frac{1.6}{0.05175} = 30.918$$ mol/kg

$$\Delta T_b = K_b \times m = 0.52 \times 30.918 = 16.08$$ °C

Boiling point = 100 + 16.08 ≈ 116 °C

The answer is $$\boxed{116}$$ °C.

Given: Solid Pb(NO₃)₂ dissolved in 1 litre of water boils at 100.15 °C. Then 0.2 mol NaCl is added, and the solution freezes at −0.8 °C.

$$K_b = 0.5$$ K kg mol$$^{-1}$$, $$K_f = 1.8$$ K kg mol$$^{-1}$$. Molality ≈ Molarity.

Pb(NO₃)₂ → Pb²⁺ + 2NO₃⁻ (van't Hoff factor $$i = 3$$)

$$m = 0.1$$ mol/kg (≈ 0.1 M)

Initial moles: Pb²⁺ = 0.1, NO₃⁻ = 0.2

NaCl → Na⁺ + Cl⁻ (complete dissociation)

Initial species: Pb²⁺ = 0.1, NO₃⁻ = 0.2, Na⁺ = 0.2, Cl⁻ = 0.2

Reaction: Pb²⁺ + 2Cl⁻ → PbCl₂↓

Let $$x$$ mol PbCl₂ precipitate. After equilibrium:

Pb²⁺ = $$(0.1 - x)$$, Cl⁻ = $$(0.2 - 2x)$$, Na⁺ = 0.2, NO₃⁻ = 0.2

Total particle molality = $$0.2 + 0.2 + (0.1-x) + (0.2-2x) = 0.7 - 3x$$

To the nearest integer: the value is $$\mathbf{13} \times 10^{-6}$$.

We need to identify how many of the given colloidal systems have liquid as the dispersion medium.

First we classify each colloidal system: Gem stones involve a solid dispersed in a solid, so the dispersion medium is solid (✗); paints involve a solid dispersed in a liquid, so the dispersion medium is liquid (✓); smoke involves a solid dispersed in a gas, so the dispersion medium is gas (✗); cheese involves a liquid dispersed in a solid (a gel), so the dispersion medium is solid (✗); milk is a liquid dispersed in a liquid (an emulsion), so the dispersion medium is liquid (✓); hair cream involves liquid or solid dispersed in a liquid (an emulsion or sol), so the dispersion medium is liquid (✓); insecticide sprays involve a liquid dispersed in a gas (an aerosol), so the dispersion medium is gas (✗); froth consists of gas dispersed in a liquid (a foam), so the dispersion medium is liquid (✓); and soap lather consists of gas dispersed in a liquid (a foam), so the dispersion medium is liquid (✓).

The colloidal systems with liquid as the dispersion medium are paints, milk, hair cream, froth, and soap lather.

Therefore, the number of colloidal systems with a liquid dispersion medium is $$5$$.

Let us examine each statement about adsorption:

Statement A: "Water vapours are adsorbed by anhydrous calcium chloride." This is incorrect. Anhydrous $$CaCl_2$$ actually absorbs water vapour (it takes water into its bulk to form hydrated salts like $$CaCl_2 \cdot 6H_2O$$), rather than adsorbing it on the surface. Adsorption is a surface phenomenon, while what $$CaCl_2$$ does is absorption — a bulk phenomenon.

Statement B: "There is a decrease in surface energy during adsorption." This is correct. Adsorption occurs because surface molecules have unbalanced forces. When adsorbate molecules attach to the surface, these residual forces are satisfied, lowering the surface energy.

Statement C: "As the adsorption proceeds, $$\Delta H$$ becomes more and more negative." This is correct. Adsorption is exothermic, and as the process continues, the cumulative enthalpy change becomes increasingly negative.

Statement D: "Adsorption is accompanied by decrease in entropy of the system." This is correct. Gas molecules lose their freedom of movement when they become bound to a surface, resulting in decreased randomness and hence decreased entropy.

Only Statement A is incorrect. The number of incorrect statements is $$\boxed{1}$$.

Hence, Correct answer is 2

We have a $$250 \text{ g}$$ solution of D-glucose in water containing $$10.8\%$$ carbon by weight. We need to find the molality.

The mass of carbon present is given by the percentage times the total mass: $$ \text{Mass of carbon} = 10.8\% \times 250 = \frac{10.8}{100} \times 250 = 27 \text{ g} $$.

The molecular formula of D-glucose is $$C_6H_{12}O_6$$, and its molecular weight is calculated as $$6(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180 \text{ g/mol}$$. Since there are 6 carbon atoms in one mole of glucose, the mass of carbon per mole is $$6 \times 12 = 72 \text{ g}$$, giving a mass fraction of carbon in glucose of $$\dfrac{72}{180} = \dfrac{2}{5}$$. Therefore, the mass of glucose in the solution is found by dividing the mass of carbon by this fraction: $$ \text{Mass of glucose} = \frac{\text{Mass of carbon}}{\text{Mass fraction of C}} = \frac{27}{2/5} = 27 \times \frac{5}{2} = 67.5 \text{ g} $$.

The mass of the solvent (water) is the difference between the total mass and the mass of glucose: $$ \text{Mass of water} = 250 - 67.5 = 182.5 \text{ g} = 0.1825 \text{ kg} $$.

The number of moles of glucose present is calculated as $$ n = \frac{67.5}{180} = 0.375 \text{ mol} $$.

The molality of the solution is then the moles of solute per kilogram of solvent: $$ m = \frac{n}{\text{mass of solvent in kg}} = \frac{0.375}{0.1825} = 2.055 \approx 2.06 $$.

The molality is nearest to $$2.06$$. The correct answer is Option B: $$2.06$$.

We need to evaluate the assertion that activated charcoal adsorbs $$SO_2$$ more efficiently than $$CH_4$$, and the reason that gases with lower critical temperatures are readily adsorbed by activated charcoal.

The ease of adsorption of a gas depends on how easily it can be liquefied, and gases with higher critical temperatures are more easily liquefiable and hence are more readily adsorbed. The critical temperature of $$SO_2$$ is $$430 \text{ K}$$ (high, easily liquefiable), while that of $$CH_4$$ is $$190 \text{ K}$$ (low, not easily liquefiable). Since $$SO_2$$ has a higher critical temperature, it is adsorbed more efficiently than $$CH_4$$, making the assertion correct.

By contrast, the reason states that gases with lower critical temperatures are readily adsorbed, which is the opposite of the correct principle. Gases with higher critical temperatures are more readily adsorbed because they are more easily liquefiable, so the reason is incorrect.

The correct answer is Option C: A is correct but R is not correct.

We need to identify which property of colloids the Zeta potential is related to.

Understanding Zeta Potential:

Zeta potential (also called electrokinetic potential) is the electric potential at the boundary of the double layer surrounding a colloidal particle. It is the potential difference between the dispersion medium and the stationary layer of fluid attached to the dispersed particle.

Evaluating the Options:

Option A: Colour

The colour of colloids is related to the scattering of light (Tyndall effect) and particle size, not to the zeta potential. Incorrect.

Option B: Brownian movement

Brownian movement is caused by the random bombardment of colloidal particles by molecules of the dispersion medium. It is related to thermal energy and particle size, not to zeta potential. Incorrect.

Option C: Charge on surface of colloidal particle

Zeta potential is directly related to the charge on the surface of colloidal particles. It measures the magnitude of the electrostatic charge at the electrical double layer around the particle. A higher zeta potential indicates greater stability of the colloidal solution due to stronger electrostatic repulsion between particles. This is correct.

Option D: Tyndall effect

The Tyndall effect is the scattering of light by colloidal particles and is related to particle size and the wavelength of light, not to zeta potential. Incorrect.

Hence, the correct answer is Option C: Charge on surface of colloidal particle.

The vapour-pressure lowering for an ideal dilute solution obeys Raoult’s law:
$$\frac{\Delta P}{P^{\circ}} = i\,X_{\text{solute}}$$
where $$\Delta P = P^{\circ}-P$$, $$P^{\circ}$$ is the vapour pressure of the pure solvent, $$P$$ is that of the solution, $$X_{\text{solute}}$$ is the mole-fraction of the solute and $$i$$ is the van’t Hoff factor.

Step 1: Calculate $$\Delta P$$ and the relative lowering
$$\Delta P = 60.000\;\text{mm Hg} - 59.724\;\text{mm Hg} = 0.276\;\text{mm Hg}$$
$$\frac{\Delta P}{P^{\circ}} = \frac{0.276}{60.000} = 4.60 \times 10^{-3}$$

Step 2: Mole fraction of the solute
Moles of solute (given) $$= 0.1\;\text{mol}$$
Mass of water $$= 1.8\;\text{kg} = 1800\;\text{g}$$
Moles of water $$= \frac{1800}{18} = 100\;\text{mol}$$
Total moles $$= 0.1 + 100 = 100.1$$
$$X_{\text{solute}} = \frac{0.1}{100.1} \approx 9.99 \times 10^{-4} \;(\text{≈}\, 1.00 \times 10^{-3})$$

Step 3: Determine the van’t Hoff factor $$i$$
$$i = \frac{\Delta P / P^{\circ}}{X_{\text{solute}}} = \frac{4.60 \times 10^{-3}}{1.00 \times 10^{-3}} = 4.6$$

Step 4: Relate $$i$$ to degree of dissociation
If one formula unit of the salt gives $$\nu$$ ions on complete dissociation, then for partial dissociation with degree $$\alpha$$:
$$i = 1 + (\nu - 1)\alpha \quad -(1)$$

The salt is 90 % dissociated, so $$\alpha = 0.90$$. Substituting $$i = 4.6$$ into equation $$(1)$$:
$$4.6 = 1 + (\nu - 1)(0.90)$$
$$4.6 - 1 = 0.90(\nu - 1)$$
$$3.6 = 0.90(\nu - 1)$$
$$\nu - 1 = \frac{3.6}{0.90} = 4$$
$$\nu = 5$$

Step 5: Final result
Each formula unit of the ionic salt produces 5 ions in solution.

Answer: 5

We have a 5% (w/v) NaCl solution to which egg albumin is added and stirred. We need to determine the type of colloidal system formed.

Albumin is a protein — a naturally occurring macromolecule. When proteins are dissolved or dispersed in an aqueous medium, they form colloidal solutions. The key question is whether this colloidal sol is lyophilic or lyophobic.

A lyophilic sol (solvent-loving) is one in which the dispersed phase has a strong affinity for the dispersion medium. Proteins like albumin have numerous polar groups ($$-NH_2$$, $$-COOH$$, $$-OH$$) on their surface that readily interact with water molecules through hydrogen bonding. Because of this strong interaction, protein sols form spontaneously when stirred — no special methods are needed, and the resulting sol is quite stable.

A lyophobic sol (solvent-hating), in contrast, is formed by substances like metals or metal sulphides that have no affinity for water. Such sols require special techniques (like Bredig's arc method) to prepare and are inherently unstable.