JEE Solutions Questions

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 ($$C_6H_5OH + C_6H_5NH_2$$)

The acid dissociates according to $$HA \; \rightleftharpoons \; H^{+} + A^{-}$$

For a colligative-property experiment we use the van’t Hoff factor $$i$$.
The freezing-point depression is related to $$i$$ by the formula
$$\Delta T_f = i \, K_f \, m$$ $$-(1)$$

Data given: $$\Delta T_f = 0.20^{\circ}\text{C}$$, $$K_f = 1.8 \text{ K kg mol}^{-1}$$, molality $$m = 0.1 \text{ m}$$.
Substituting in $$(1)$$:

$$0.20 = i \,(1.8)\,(0.1)$$

$$\Rightarrow \; i = \frac{0.20}{1.8 \times 0.1} = \frac{0.20}{0.18} = 1.11$$ (keep three significant figures)

For a weak monobasic acid each formula unit can produce two ions, so $$n = 2$$.
The relation between $$i$$ and the degree of dissociation $$\alpha$$ is
$$i = 1 + \alpha (n-1) = 1 + \alpha$$ $$-(2)$$

Using $$i = 1.11$$ in $$(2)$$:
$$\alpha = i - 1 = 1.11 - 1 = 0.11$$

The equilibrium concentration of the acid is $$c = 0.1 \text{ mol L}^{-1}$$.
For a weak monoprotic acid the dissociation constant is
$$K_a = \frac{c \alpha^{2}}{1 - \alpha}$$ $$-(3)$$

Substituting the values:
$$K_a = \frac{0.1 \times (0.11)^{2}}{1 - 0.11} = \frac{0.1 \times 0.0121}{0.89} = \frac{0.00121}{0.89} \approx 1.4 \times 10^{-3}$$

Therefore $$K_a \approx 1.38 \times 10^{-3}$$.

Option A is correct.

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

We use the colligative property of freezing point depression to determine the van’t Hoff factor $$i$$.

The formula for depression in freezing point is $$\Delta T_f = i \cdot K_f \cdot m$$.

Given: $$\Delta T_f = 0.558°C$$, $$K_f = 1.86$$ K kg mol^{-1}, $$m = 0.1$$ molal.

Substituting these values yields $$0.558 = i \times 1.86 \times 0.1$$ so that $$i = \frac{0.558}{0.186} = 3$$.

A van’t Hoff factor of $$3$$ indicates that the complex dissociates into three ions in solution.

Complex $$[Cr(NH_3)_5Cl]Cl_2$$ dissociates into $$[Cr(NH_3)_5Cl]^{2+}$$ and $$2Cl^-$$, giving three ions (i = 3).

Complex $$[Cr(NH_3)_6Cl]Cl_3$$ would involve a coordination number of seven (which is not possible for Cr since CN = 6).

Complex $$[Cr(NH_3)_3Cl_3]$$ is a neutral species and does not produce free ions (i = 1).

Complex $$[Cr(NH_3)_4Cl_2]Cl$$ dissociates into two ions (i = 2).

Since chromium must have coordination number six and the van’t Hoff factor is three, the only viable structure is $$[Cr(NH_3)_5Cl]Cl_2$$.

The correct answer is Option 1: $$[Cr(NH_3)_5Cl]Cl_2$$.

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).

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 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.

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 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): $$(CH_3)_2CO + CS_2$$.

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

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.

Given: Molarity = 0.2 M, volume = 500 mL, density = 1.25 g/mL.

Moles of $$CuSO_4$$ = $$0.2 \times 0.5 = 0.1$$ mol.

Molar mass of anhydrous $$CuSO_4$$ = 64 + 32 + 64 = 160 g/mol.

Mass of $$CuSO_4$$ = $$0.1 \times 160 = 16$$ g.

Mass of solution = $$500 \times 1.25 = 625$$ g.

Mass of solvent (water) = $$625 - 16 = 609$$ g = 0.609 kg.

Molality = $$\frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.1}{0.609} = 0.16420$$ m.

Converting: $$0.16420$$ m = $$164.2 \times 10^{-3}$$ m.

Nearest integer: $$164 \times 10^{-3}$$ m.

The answer is $$\boxed{164}$$.

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 degree of ionization $$\alpha$$ for electrolyte $$AB_2$$.

Boiling point elevation is calculated as $$\Delta T_b = 100.52 - 100 = 0.52°C$$.

Since $$\Delta T_b = iK_bm$$, where $$m$$ is the molality, we first determine $$m$$.

Substituting the values gives $$m = \frac{10/200}{100/1000} = \frac{0.05}{0.1} = 0.5 \text{ mol/kg}$$.

Using $$\Delta T_b = iK_bm$$ again with $$0.52 = i \times 0.52 \times 0.5$$ leads to $$i = \frac{0.52}{0.26} = 2$$.

The dissociation of $$AB_2$$ is represented by $$AB_2 \rightarrow A^{2+} + 2B^-$$, yielding $$n = 3$$ ions.

Relating the van't Hoff factor to the degree of ionization via $$i = 1 + (n-1)\alpha$$, we have $$2 = 1 + 2\alpha$$, which gives $$\alpha = 0.5$$.

Therefore, $$\alpha = 0.5 = 5 \times 10^{-1}$$.

The correct answer is 5.

An artificial cell containing 0.2 M glucose is placed in a 0.05 M NaCl solution (complete dissociation yields 0.1 M total ion concentration). The osmotic pressure can be calculated using the formula $$\pi = iCRT$$, where $$i$$ is the van ’t Hoff factor, $$C$$ the molar concentration, $$R$$ the gas constant, and $$T$$ the temperature.

For the glucose solution (a non-electrolyte with $$i = 1$$), the osmotic pressure inside the cell is $$\pi_{inside} = 1 \times 0.2 \times R \times T$$. For the NaCl solution (with $$i = 2$$ under complete dissociation), the osmotic pressure outside the cell is $$\pi_{outside} = 2 \times 0.05 \times R \times T = 0.1 \times R \times T$$.

The net osmotic pressure difference is given by $$\pi = \pi_{inside} - \pi_{outside} = (0.2 - 0.1) \times R \times T = 0.1 \times 0.083 \times 300$$ which evaluates to $$2.49 \text{ bar} = 24.9 \times 10^{-1} \text{ bar} \approx 25 \times 10^{-1} \text{ bar}$$.

The final answer is 25.

We need to find the osmotic pressure of a 0.03 M aqueous solution of weak acid HX. The acid dissociates as $$HX \rightleftharpoons H^+ + X^-$$, and we are given $$K_a = 1.2 \times 10^{-5}$$, $$C = 0.03$$ M, $$T = 300$$ K, and $$R = 0.083$$ L bar mol$$^{-1}$$ K$$^{-1}$$.

We begin by finding the degree of dissociation, $$\alpha$$. At equilibrium, $$[HX] = C(1-\alpha)$$, $$[H^+] = C\alpha$$, and $$[X^-] = C\alpha$$, so

$$K_a = \frac{C\alpha \times C\alpha}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha}$$

For small $$\alpha$$ ($$\alpha \ll 1$$), we approximate $$K_a \approx C\alpha^2$$, which gives

$$\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.2 \times 10^{-5}}{0.03}} = \sqrt{4 \times 10^{-4}} = 0.02$$

Since $$\alpha = 0.02$$ (2%), the approximation $$\alpha \ll 1$$ is valid.

The van’t Hoff factor, $$i$$, for a substance that dissociates into 2 ions with degree of dissociation $$\alpha$$ is

$$i = 1 + \alpha(n - 1) = 1 + \alpha(2 - 1) = 1 + \alpha = 1 + 0.02 = 1.02$$

Finally, the osmotic pressure is given by

$$\pi = iCRT$$

Substituting the values,

$$\pi = 1.02 \times 0.03 \times 0.083 \times 300$$

First compute $$CRT$$:

$$CRT = 0.03 \times 0.083 \times 300 = 0.747$$ bar

Thus,

$$\pi = 1.02 \times 0.747 = 0.76194$$ bar

$$= 76.194 \times 10^{-2}$$ bar $$\approx 76 \times 10^{-2}$$ bar

The answer is 76.

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 = $$707\ \text{mm Hg}$$.

We need to find the mass percent of ethyl alcohol in a solution containing 1 mole ethyl alcohol and 9 moles of water.

Since the mass percent of solute is given by $$\text{Mass percent} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100$$, we note that the molar mass of ethyl alcohol is 46 g/mol and that of water is 18 g/mol.

Now the mass of ethyl alcohol in the solution is $$1 \times 46 = 46$$ g, and the mass of water is $$9 \times 18 = 162$$ g. Substituting these values, the total mass of the solution becomes $$46 + 162 = 208$$ g.

Therefore the mass percent of ethyl alcohol is $$\text{Mass percent} = \frac{46}{208} \times 100 = \frac{4600}{208} = 22.115\%$$, which rounds to approximately $$22\%$$.

The answer is $$\boxed{22}$$.

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 $$\boxed{4}$$.

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 $$\boxed{23}$$.

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 mole fraction of urea in an aqueous solution with molality 4.44 m.

Molality (m) is defined as the number of moles of solute per kilogram of solvent:

$$m = \frac{n_{\text{solute}}}{W_{\text{solvent (kg)}}}$$

For molality = 4.44 m, we have 4.44 moles of urea in 1 kg (1000 g) of water.

Moles of water = $$\frac{1000}{18} = 55.56$$ mol

The mole fraction of urea is:

$$x_{\text{urea}} = \frac{n_{\text{urea}}}{n_{\text{urea}} + n_{\text{water}}} = \frac{4.44}{4.44 + 55.56} = \frac{4.44}{60} = 0.074$$

Given that the mole fraction = $$x \times 10^{-3}$$:

$$0.074 = x \times 10^{-3}$$

$$x = \frac{0.074}{10^{-3}} = 74$$

The answer is $$\boxed{74}$$.

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 = $$\frac{2}{20} \times 100 = 10\%$$

Determine what "25 volume" hydrogen peroxide means in terms of mass of $$H_2O_2$$ per litre.

"25 volume" $$H_2O_2$$ means 1 L of this solution releases 25 L of $$O_2$$ at STP upon complete decomposition.

The decomposition reaction is $$2H_2O_2 \rightarrow 2H_2O + O_2$$.

Since 1 mole of $$H_2O_2$$ produces $$\frac{1}{2}$$ mole of $$O_2$$, which corresponds to 11.2 L at STP, the volume strength is $$11.2 \times M$$ where M is the molarity of $$H_2O_2$$. Equating this to 25 gives $$25 = 11.2 \times M$$, so $$M = \frac{25}{11.2} = 2.232$$ mol/L.

The molar mass of $$H_2O_2$$ is 34 g/mol, so the mass per litre is $$2.232 \times 34 \approx 75.9$$ g, which is approximately 75 g.

Comparing with the given options, 250 g/L (Option A) is too high; 75 g/L (Option B) matches our calculation $$\checkmark$$; 25 g per 100 mL = 250 g/L (Option C) is too high; and 25 g/L (Option D) is too low. Thus the correct answer is Option B.

For one mole of a real gas, the Van der Waals equation is

$$\left(P+\frac{a}{V^2}\right)(V-b)=RT.$$

Expanding the equation,

$$PV-Pb+\frac{a}{V}-\frac{ab}{V^2}=RT.$$

Dividing throughout by (RT),

$$\frac{PV}{RT}-\frac{Pb}{RT}+\frac{a}{RTV}-\frac{ab}{RTV^2}=1.$$

Since the compressibility factor is defined as

$$Z=\frac{PV}{RT},$$

we obtain

$$Z-\frac{Pb}{RT}+\frac{a}{RTV}-\frac{ab}{RTV^2}=1.$$

At low pressures, the molar volume (V) is very large, so terms containing (\frac{1}{V^2}) can be neglected. Therefore,

$$Z\approx 1-\frac{a}{RTV}.$$

Point A corresponds to the low-pressure region where intermolecular attractive forces dominate, causing the gas to be more compressible than an ideal gas ((Z<1)). The deviation from ideal behaviour is represented by the attractive term (-\frac{a}{RTV}).

Hence, the compressibility factor at point A is

$$\boxed{Z=1-\frac{a}{RTV}}.$$

Therefore, the correct answer is

$$\boxed{1-\frac{a}{RTV}}.$$

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)

Hydration enthalpy depends on the charge density of the ion: higher charge and smaller ionic radius lead to greater (more negative) hydration enthalpy.

The ions given are (A) K$$^+$$ (charge = +1, ionic radius = 138 pm), (B) Rb$$^+$$ (charge = +1, ionic radius = 152 pm), (C) Mg$$^{2+}$$ (charge = +2, ionic radius = 72 pm), (D) Cs$$^+$$ (charge = +1, ionic radius = 167 pm) and (E) Ca$$^{2+}$$ (charge = +2, ionic radius = 100 pm).

Divalent cations have much higher hydration enthalpies than monovalent cations due to their greater charge density.

Between the divalent ions, Mg$$^{2+}$$ is smaller than Ca$$^{2+}$$, so Mg$$^{2+}$$ has greater hydration enthalpy: C > E.

Among the monovalent ions, smaller size means greater hydration enthalpy: K$$^+$$ > Rb$$^+$$ > Cs$$^+$$, i.e., A > B > D.

The complete order is:

$$\text{Mg}^{2+} > \text{Ca}^{2+} > \text{K}^+ > \text{Rb}^+ > \text{Cs}^+$$

That is: C > E > A > B > D.

The correct answer is Option D.

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

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

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.

Osmotic pressure is given by $$\pi=\dfrac{n_{\text{total}}RT}{V_{\text{solution}}}$$, where
$$n_{\text{total}}$$ = total number of moles of solute after mixing,
$$R = 62\;{\text{L Torr K}}^{-1}{\text{mol}}^{-1}$$ (given),
$$T = 300\;\text{K}$$ (given),
$$V_{\text{solution}}$$ = final volume of the mixed solution in litres.

Step 1: Moles of urea in the 50 mL, 0.2 molal solution

Molality definition: $$m = \dfrac{n}{\text{mass of solvent (kg)}}$$

Let the mass of solvent in this solution be $$w\; \text{g}$$.
Then $$n = m \times \dfrac{w}{1000} = 0.2 \times \dfrac{w}{1000}=0.0002w$$ mol.

Mass of solute $$= n \times M = 0.0002w \times 60 = 0.012w \;\text{g}$$.

Total mass of this solution $$= w + 0.012w = 1.012w \;\text{g}$$.

Using the density (1.012 g mL$$^{-1}$$):
$$\text{mass} = \text{density} \times \text{volume} = 1.012 \times 50 = 50.6 \;\text{g}$$.

So $$1.012w = 50.6 \;\Rightarrow\; w = 50.0\;\text{g}$$.

Therefore, moles of urea in this portion
$$n_1 = 0.0002 \times 50 = 0.01\;\text{mol}$$.

Step 2: Moles of urea in the 250 mL second solution

Given mass of urea = 0.06 g.
$$n_2 = \dfrac{0.06}{60} = 0.001\;\text{mol}$$.

Step 3: Total moles after mixing

$$n_{\text{total}} = n_1 + n_2 = 0.01 + 0.001 = 0.011\;\text{mol}$$.

Step 4: Total volume of the mixed solution

Assuming $$\Delta_{mix}V = 0$$ (volume is additive):
$$V_{\text{solution}} = 50\;\text{mL} + 250\;\text{mL} = 300\;\text{mL} = 0.300\;\text{L}$$.

Step 5: Osmotic pressure

$$\pi = \dfrac{n_{\text{total}} RT}{V_{\text{solution}}} = \dfrac{0.011 \times 62 \times 300}{0.300}$$

First compute the concentration:
$$\dfrac{0.011}{0.300}=0.0366667\;\text{mol L}^{-1}$$.

Then
$$\pi = 0.0366667 \times 62 \times 300 = 0.0366667 \times 18\,600 \approx 682\;\text{Torr}$$.

Hence, the osmotic pressure of the resulting solution at 300 K is 682 Torr.

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 $$364$$.

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 $$\boxed{139}$$.

1. Understand the Chemical Equation

The decomposition of hydrogen peroxide ($$\text{H}_2\text{O}_2$$) is represented by the balanced chemical equation:

$$2\text{H}_2\text{O}_2\text{ (aq)} \rightarrow 2\text{H}_2\text{O}\text{ (l)} + \text{O}_2\text{ (g)}$$

From the stoichiometry of the reaction:

• 2 moles of $$\text{H}_2\text{O}_2$$ produce 1 mole of $$\text{O}_2$$ gas.

2. Define "Volume Strength"

A "50 volume" solution means that $$1\text{ L}$$ of this $$\text{H}_2\text{O}_2$$ solution will liberate $$50\text{ L}$$ of $$\text{O}_2$$ gas at STP.

3. Calculate Moles of $$\text{O}_2$$ Produced

Given that the molar volume of a gas at STP is $$22.7\text{ L}$$:

$$\text{Moles of }\text{O}_2 = \frac{\text{Volume of }\text{O}_2\text{ at STP}}{\text{Molar Volume}} = \frac{50\text{ L}}{22.7\text{ L/mol}}$$

4. Calculate Moles and Mass of $$\text{H}_2\text{O}_2$$

Since 1 mole of $$\text{O}_2$$ comes from 2 moles of $$\text{H}_2\text{O}_2$$:

$$\text{Moles of }\text{H}_2\text{O}_2 = 2 \times \text{Moles of }\text{O}_2 = 2 \times \frac{50}{22.7} = \frac{100}{22.7}\text{ moles}$$

Now, we can find the mass of $$\text{H}_2\text{O}_2$$ present in that $$1\text{ L}$$ solution using its molar mass ($$34\text{ g/mol}$$):

$$\text{Mass of }\text{H}_2\text{O}_2 = \text{Moles} \times \text{Molar Mass} = \frac{100}{22.7} \times 34 = \frac{3400}{22.7}\text{ g}$$
$$\text{Mass of }\text{H}_2\text{O}_2 \approx 149.78\text{ g}$$

The strength of the solution is the mass of solute per liter of solution ($$\text{g/L}$$). Since this mass is contained in $$1\text{ L}$$:

$$\text{Strength} \approx 150\text{ g/L}$$

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$$^{-1}$$.

Statement A is incorrect. Molecules present in the bulk of a liquid are surrounded by neighbouring molecules from all directions and experience equal intermolecular attractive forces, resulting in zero net force. Surface tension does not arise due to these balanced forces.

Statement B is correct. Molecules present at the surface experience unbalanced intermolecular forces because there are no molecules above them. This net inward force causes the liquid surface to contract and gives rise to surface tension.

Statement C is incorrect. Molecules in a liquid are in continuous random motion and can move freely between the bulk and the surface. Therefore, a molecule in the bulk can reach the liquid surface.

Statement D is correct. Vapour pressure is produced when surface molecules escape into the vapour phase in a closed system until dynamic equilibrium is established. Hence, the molecules present on the surface are responsible for vapour pressure.

Therefore, Statements B and D are correct, while Statements A and C are incorrect. Hence, the number of correct statements is $$\boxed{2}$$.

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.

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 $$5$$.

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.

Since albumin is a protein with strong affinity for water, the dispersion formed is a lyophilic sol. The NaCl solution simply serves as the dispersion medium — at 5% concentration, it is not high enough to salt out the albumin.

Hence, the correct answer is Option A: Lyophilic sol.

Let us analyze both the Assertion and the Reason.

Assertion: A 5 M solution of KCl at $$10°C$$ has density 'x' g/mL. When cooled to $$-21°C$$, the molality of the solution remains unchanged.

Reason: Molality does not change with temperature as mass remains unaffected with temperature.

Checking the Assertion: Molality is defined as the number of moles of solute per kilogram of solvent:

$$m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}}$$

When the temperature changes, the mass of the solute and the mass of the solvent remain the same (mass does not depend on temperature). Therefore, molality does not change with temperature. The assertion is true.

Checking the Reason: The reason states that molality does not change with temperature because mass remains unaffected by temperature. This is exactly why molality is temperature-independent — unlike molarity which depends on volume (which changes with temperature), molality depends only on mass. The reason is true and correctly explains the assertion.

Both the Assertion and Reason are true, and the Reason is the correct explanation of the Assertion.

Hence, the correct answer is Option C.

We need to evaluate two statements about emulsions.

Statement I: "Emulsions of oil in water are unstable and sometimes they separate into two layers on standing."

This statement is correct. Emulsions are thermodynamically unstable systems. Without an emulsifying agent (stabilizer), oil-in-water emulsions tend to separate into two distinct layers over time due to the tendency to minimize the interfacial area between the two immiscible liquids. This process is called demulsification or creaming.

Statement II: "For stabilisation of an emulsion, excess of electrolyte is added."

This statement is incorrect. Emulsions are stabilized by adding emulsifying agents (also called surfactants or emulsifiers) such as soaps, detergents, proteins (like casein), or gums. These substances form a protective layer around the dispersed droplets, preventing coalescence.

In fact, adding excess electrolyte to an emulsion can actually destabilize it by compressing the electrical double layer around the dispersed droplets (similar to coagulation of colloids). This is the opposite of stabilization.

Therefore, Statement I is correct but Statement II is incorrect.

The correct answer is Option C.

We are given that solute A associates in water and we need to find the percentage association.

Since the freezing point depression is given by $$\Delta T_f = K_f \times m$$, substituting $$\Delta T_f = 0.2$$ and $$K_f = 1.86$$ yields $$m = \frac{0.2}{1.86} = 0.1075 \text{ mol/kg}$$.

Without association, the theoretical molality is calculated using the mass of solute and solvent; thus, $$m_{\text{theoretical}} = \frac{0.7}{93 \times 0.042} = \frac{0.7}{3.906} = 0.1792 \text{ mol/kg}$$.

From the ratio of observed to theoretical molality, the van't Hoff factor is $$i = \frac{m_{\text{observed}}}{m_{\text{theoretical}}} = \frac{0.1075}{0.1792} = 0.6$$.

For a dimerization association where $$n = 2$$, the relation between $$i$$ and the fraction associated $$x$$ is $$i = 1 - x + \frac{x}{n} = 1 - x + \frac{x}{2} = 1 - \frac{x}{2}$$. Substituting $$i = 0.6$$ gives $$0.6 = 1 - \frac{x}{2}$$, which leads to $$\frac{x}{2} = 0.4$$ and hence $$x = 0.8$$ or 80\%.

Therefore, the percentage association of solute A in water is 80%, corresponding to Option D.

We need to evaluate the Assertion and Reason about dialysis of colloidal solutions.

Analyzing Assertion A:

Assertion A states that dissolved substances can be removed from a colloidal solution by diffusion through a parchment paper. This is the principle of dialysis. In dialysis, a colloidal solution is placed in a bag made of parchment paper (a semi-permeable membrane). The dissolved ions and molecules (true solution particles) pass through the parchment paper into the outer liquid, while the larger colloidal particles are retained.

$$\therefore$$ Assertion A is correct.

Analyzing Reason R:

Reason R states that particles in a true solution cannot pass through parchment paper but colloidal particles can. This is the opposite of what actually happens:

- True solution particles (ions, small molecules) can pass through parchment paper.

- Colloidal particles are too large and cannot pass through parchment paper.

$$\therefore$$ Reason R is incorrect.

Conclusion:

A is correct but R is not correct.

Hence, the correct answer is Option C.

The Tyndall effect is the scattering of light by colloidal particles. We need to identify the incorrect statement about it.

Recall conditions for the Tyndall effect: The Tyndall effect occurs when:

$$\bullet$$ The refractive indices of the dispersed phase and dispersion medium differ significantly.

$$\bullet$$ The diameter of the dispersed particles is not too small compared to the wavelength of light used — typically in the range of 1 nm to 1000 nm (comparable to the wavelength of visible light, 400-700 nm).

Analyze each statement: Option A: "The refractive indices of the dispersed phase and the dispersion medium differ greatly in magnitude." — This is correct. A large difference in refractive index enhances scattering.

Option B: "The diameter of the dispersed particles is much smaller than the wavelength of the light used." — This is incorrect. If the particles were much smaller than the wavelength (as in true solutions), they would not scatter light effectively. For the Tyndall effect, the particle size must be comparable to or not too much smaller than the wavelength of light.

Option C: "It is used to distinguish a true solution from a colloidal solution." — This is correct. True solutions do not show the Tyndall effect, while colloidal solutions do.

Option D: "During projection of movies in the cinema hall, Tyndall effect is noticed." — This is correct. The visible beam of light from the projector passing through dusty air is an example of the Tyndall effect.

The correct answer is Option B: The diameter of the dispersed particles is much smaller than the wavelength of the light used.

Let us match each item in List-I with the correct item in List-II.

A. Lyophilic colloid → II (Protective colloid)

Lyophilic colloids (solvent-loving) are stable colloids that can protect lyophobic colloids from coagulation. They are also called protective colloids. Examples include gelatin, starch, and gum arabic.

B. Emulsion → I (Liquid-liquid colloid)

An emulsion is a colloidal system where both the dispersed phase and the dispersion medium are liquids. For example, milk is an emulsion of fat droplets in water. Hence, emulsion is a liquid-liquid colloid.

C. Positively charged colloid → IV ($$FeCl_3$$ + hot water)

When $$FeCl_3$$ is added to hot water, it hydrolyzes to form a colloidal solution of $$Fe(OH)_3$$ which is positively charged. The $$Fe^{3+}$$ ions are preferentially adsorbed on the $$Fe(OH)_3$$ particles, giving them a positive charge.

D. Negatively charged colloid → III ($$FeCl_3 + NaOH$$)

When $$FeCl_3$$ is mixed with excess $$NaOH$$, the $$Fe(OH)_3$$ sol formed adsorbs $$OH^-$$ ions from the excess NaOH, making the colloidal particles negatively charged.

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

Hence, the correct answer is Option A.

We need to match each item in List-I with the correct item in List-II based on properties of colloids.

CdS sol is a well-known example of a negatively charged sol. Other examples include As$$_2$$S$_3$$ sol and gold sol. So (A) negatively charged sol → (II) CdS sol.

Starch is a macromolecular colloid — it is a naturally occurring polymer that forms colloidal solutions in water. So (B) macromolecular colloid → (III) Starch.

Fe$$_2$$O$$_3$$ · xH$$_2$$O (ferric oxide hydrosol) is a positively charged sol. Other examples include Al(OH)$$_3$$ sol. So (C) positively charged sol → (I) Fe$$_2$$O$$_3$$ · xH$$_2$$O.

Cheese is a well-known example of a gel (solid dispersed in liquid medium that has a semi-solid structure). So (D) Cheese → (IV) a gel.

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

Hence, the correct answer is Option C.

We are given two solutions A and B, each prepared by dissolving 1 g of non-volatile solutes X and Y respectively in 1 kg of water, and the ratio of depression in freezing points is 1 : 4.

Starting with the formula for freezing point depression, we have

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

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

Since 1 g of solute is dissolved in 1 kg of solvent, the molality can be expressed in terms of the molar mass $$M$$ as

$$m = \frac{1}{M \times 1} = \frac{1}{M}$$

Here, $$M$$ represents the molar mass of the solute (in g/mol).

When comparing the freezing point depressions of solutions A and B, we obtain

$$\frac{\Delta T_{f,A}}{\Delta T_{f,B}} = \frac{K_f / M_X}{K_f / M_Y} = \frac{M_Y}{M_X}$$

Given this ratio equals $$\frac{1}{4}$$, it follows that

$$\frac{M_Y}{M_X} = \frac{1}{4}$$

Hence,

$$\frac{M_X}{M_Y} = \frac{4}{1} = \frac{1}{0.25}$$

This establishes the molar mass ratio $$M_X : M_Y = 1 : 0.25$$, implying $$M_X = 4 \times M_Y$$.

Therefore, the correct choice is Option B.

Given:

$$42.12\% \text{ (w/v) solution of NaCl}$$

This means:

$$42.12\,\text{g NaCl present in }100\,\text{mL solution}$$

Molar mass of NaCl:

$$=23+35.5$$

$$=58.5\,\text{g mol}^{-1}$$

Number of moles of NaCl:

$$=\frac{42.12}{58.5}$$

$$=0.72\,\text{mol}$$

Volume of solution:

$$100\,\text{mL}=0.1\,\text{L}$$

Hence molarity of NaCl solution:

$$=\frac{0.72}{0.1}$$

$$=7.2\,\text{mol L}^{-1}$$

The solution causes coagulation in $$10$$ hours.

Coagulating value is expressed in millimoles per litre:

$$\text{Coagulating value}=\frac{7.2 \times 1000}{5}$$

$$=1440\,\text{mmol L}^{-1}$$

Final Answer:

$${1440\,\text{mmol L}^{-1}}$$

Hence, Option D is correct

We are given that the boiling point of a 2% aqueous solution of a non-volatile solute A is equal to the boiling point of an 8% aqueous solution of a non-volatile solute B. We need to find the relation between their molecular weights $$M_A$$ and $$M_B$$.

Recall the boiling point elevation formula. The elevation in boiling point for a dilute solution is given by:

$$ \Delta T_b = K_b \times m $$

where $$K_b$$ is the ebullioscopic constant of the solvent and $$m$$ is the molality of the solution.

Since both solutions are aqueous, they share the same solvent (water) and hence the same $$K_b$$. Equal boiling points mean equal boiling point elevations:

$$ \Delta T_{b,A} = \Delta T_{b,B} $$

$$ K_b \times m_A = K_b \times m_B $$

Cancelling $$K_b$$ gives

$$ m_A = m_B $$

A "w%" aqueous solution means w grams of solute is dissolved in 100 grams of solution, so the mass of solvent (water) is (100 - w) grams. Molality is defined as moles of solute per kilogram of solvent:

$$ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{w / M}{\frac{(100 - w)}{1000}} = \frac{1000w}{M \times (100 - w)} $$

For 2% solution of A: $$ m_A = \frac{1000 \times 2}{M_A \times (100 - 2)} = \frac{2000}{98 \, M_A} $$

For 8% solution of B: $$ m_B = \frac{1000 \times 8}{M_B \times (100 - 8)} = \frac{8000}{92 \, M_B} $$

Equating these molalities and cross-multiplying yields:

$$ \frac{2000}{98 \, M_A} = \frac{8000}{92 \, M_B} $$

$$ 2000 \times 92 \times M_B = 8000 \times 98 \times M_A $$

$$ 184000 \, M_B = 784000 \, M_A $$

Dividing both sides by 184000 gives

$$ M_B = \frac{784000}{184000} \, M_A = \frac{784}{184} \, M_A $$

Simplifying the fraction:

$$ \frac{784}{184} = \frac{784 \div 8}{184 \div 8} = \frac{98}{23} \approx 4.26 $$

Applying the standard JEE approximation for dilute solutions (where the solute percentage is small) under which (100 - w) ≈ 100, we have:

$$ m_A = \frac{1000 \times 2}{M_A \times 100} = \frac{20}{M_A} $$

$$ m_B = \frac{1000 \times 8}{M_B \times 100} = \frac{80}{M_B} $$

Setting these equal,

$$ \frac{20}{M_A} = \frac{80}{M_B} $$

$$ 20 \, M_B = 80 \, M_A $$

$$ M_B = \frac{80}{20} \, M_A = 4 \, M_A $$

Therefore, $$M_B = 4M_A$$.

The correct answer is Option B: $$M_B = 4M_A$$.

We are given the critical temperatures of four gases — He (5.2 K), $$CH_4$$ (190 K), $$CO_2$$ (304.2 K), and $$NH_3$$ (405.5 K) — and asked which gas shows the least adsorption on charcoal.

Adsorption of a gas on a solid surface depends on how easily the gas can be liquefied. A gas that is more easily liquefied (i.e., has stronger intermolecular forces) will be adsorbed more readily on the surface of charcoal. The critical temperature of a gas is a direct measure of the strength of its intermolecular forces: the higher the critical temperature, the stronger the intermolecular attractions, and the more easily the gas can be liquefied.

Now, among the given gases, $$NH_3$$ has the highest critical temperature (405.5 K), meaning it has the strongest intermolecular forces (due to hydrogen bonding) and is most easily liquefied — so it will be adsorbed the most. $$CO_2$$ (304.2 K) and $$CH_4$$ (190 K) follow in decreasing order of adsorption.

Helium has the lowest critical temperature of just 5.2 K, indicating extremely weak van der Waals forces. It is the hardest gas to liquefy among the four, and therefore it will show the least adsorption on charcoal.

Hence, the correct answer is Option A.

We need to find the Van't Hoff factor (i) for formic acid solution.

Concentration = $$0.5 \text{ mL L}^{-1}$$ means 0.5 mL of formic acid in 1 L of solution (approximately 1 L of water for a dilute solution), so the mass of formic acid is Volume $$\times$$ Density = $$0.5 \times 1.05 = 0.525 \text{ g}$$.

The molar mass of formic acid ($$HCOOH$$) is $$1 + 12 + 16 + 16 + 1 = 46 \text{ g mol}^{-1}$$, and the moles of formic acid present are $$\dfrac{0.525}{46} = 0.01141 \text{ mol}$$. Since the solution is very dilute, the mass of the solvent (water) is approximately 1000 g = 1 kg, giving a molality of $$\dfrac{0.01141}{1} = 0.01141 \text{ mol kg}^{-1}$$.

Applying the depression in freezing point formula $$\Delta T_f = i \cdot K_f \cdot m$$, we have $$0.0405 = i \times 1.86 \times 0.01141$$ which simplifies to $$0.0405 = i \times 0.02122$$, and hence $$i = \dfrac{0.0405}{0.02122} = 1.908 \approx 1.9$$.

The Van't Hoff factor $$i \approx 1.9$$ indicates that formic acid nearly completely dissociates in water as: $$HCOOH \rightleftharpoons HCOO^- + H^+$$, giving approximately 2 particles per molecule.

Hence, the correct answer is Option C.

We need to identify the correct statement about colloidal solutions. Option A, “Brownian motion destabilises sols,” is incorrect because Brownian motion actually stabilises sols by preventing the colloidal particles from settling down under gravity; the continuous random motion keeps the particles suspended. Option B, “Any amount of dispersed phase can be added to emulsion without destabilising it,” is incorrect because adding too much dispersed phase can cause phase inversion or destabilisation since emulsions have a limited capacity to hold dispersed phase. Option C, “Mixing two oppositely charged sols in equal amount neutralises charges and stabilises colloids,” is incorrect as mixing oppositely charged sols causes mutual coagulation (precipitation), not stabilisation; the opposite charges neutralise each other, removing the electrostatic stabilisation, and the particles aggregate and settle.

Option D, “Presence of equal and similar charges on colloidal particles provides stability to the colloidal solution,” is correct because when all colloidal particles carry the same type of charge (either all positive or all negative), they repel each other. This electrostatic repulsion prevents the particles from coming together and coagulating, thus providing stability to the colloidal solution. This is known as electrostatic stabilisation.

The correct answer is Option D.

We need to determine which statements about micelle formation are correct.

Micelle formation occurs when surfactant molecules aggregate in solution above the critical micelle concentration (CMC). This process involves the hydrophobic tails coming together, releasing ordered water molecules from around the hydrophobic parts.

Since micelle formation involves breaking the structured water cage around the hydrophobic tails, it is generally an endothermic process ($$\Delta H > 0$$). Energy is required to break the structured water cage (iceberg structure) around the hydrophobic tails, so statement (B) is correct and (A) is incorrect.

Upon micelle formation, the ordered water molecules surrounding the hydrophobic parts of the surfactant are released into the bulk water. This release leads to a large positive entropy change ($$\Delta S > 0$$), so statement (C) is correct and (D) is incorrect.

Micelle formation is spontaneous ($$\Delta G < 0$$) because the large positive $$T\Delta S$$ term overcomes the positive $$\Delta H$$:

$$\Delta G = \Delta H - T\Delta S < 0$$

This confirms that the process is entropy-driven.

Therefore, statements B and C are correct.

Hence, the correct choice is Option C.

We need to evaluate the assertion and reason about colloidal gold.

Assertion A: Finest gold is red in colour, as the size of the particles increases, it appears purple then blue and finally gold.

This is correct. Colloidal gold nanoparticles exhibit different colours depending on their size. Very fine particles ($$\sim$$20 nm) appear red, medium-sized particles appear purple/blue, and bulk gold has its characteristic golden colour.

Reason R: The colour of the colloidal solution depends on the wavelength of light scattered by the dispersed particles.

This is correct. The colour of colloidal solutions depends on the wavelength of scattered light. As particle size changes, the wavelength of light scattered changes, which results in different observed colours. This phenomenon is related to the Tyndall effect and surface plasmon resonance in metal nanoparticles.

Is R the correct explanation of A?

Yes. The reason gold colloids of different sizes appear as different colours (Assertion A) is precisely because the wavelength of scattered light depends on the particle size (Reason R). Smaller particles scatter shorter wavelengths, while larger particles scatter longer wavelengths, producing the observed colour changes from red to purple to blue to gold.

Therefore, the correct answer is Option A: Both A and R are true and R is the correct explanation of A.

This question asks why using very little soap is ineffective for cleaning clothes.

Soap molecules are amphiphilic — they have a hydrophobic (non-polar) tail and a hydrophilic (polar) head. For effective cleaning, they must form micelles — spherical aggregates where the hydrophobic tails point inward (trapping grease) and the hydrophilic heads face outward (toward water). However, micelles form only when the concentration of soap in water exceeds a certain threshold called the Critical Micelle Concentration (CMC); below this concentration, soap molecules remain as individual ions or molecules and cannot aggregate into micelles.

When very little soap is used, the soap concentration in water remains below the CMC, preventing micelle formation and thus rendering the soap unable to effectively trap and remove grease and dirt from clothes.

Option A: Soap particles remain floating in water as ions — this is partially true but does not explain why cleaning fails.
Option B: Colloidal structure of soap in water is completely disturbed — incorrect, as the colloidal structure (micelles) was never formed in the first place.
Option C: The hydrophobic part of soap is not able to take away grease — incorrect reasoning; the hydrophobic part can still interact with grease, but without micelle formation, it cannot be effectively removed.
Option D: The micelles are not formed due to concentration of soap below its CMC value — this is the precise scientific explanation.

The correct answer is Option D: The micelles are not formed due to concentration of soap below its CMC value.

We need to find the molarity of the remaining nitric acid solution after heating.

Initial volume = $$800 \text{ mL} = 0.8 \text{ L}$$

Initial molarity = $$0.5 \text{ M}$$

$$\text{Initial moles of } HNO_3 = 0.5 \times 0.8 = 0.4 \text{ mol}$$

Mass of nitric acid evaporated = $$11.5 \text{ g}$$

Molar mass of $$HNO_3$$ = $$63 \text{ g/mol}$$

$$\text{Moles evaporated} = \frac{11.5}{63} = \frac{11.5}{63} \approx 0.1825 \text{ mol}$$

$$\text{Remaining moles} = 0.4 - 0.1825 = 0.2175 \text{ mol}$$

The volume is reduced to half:

$$\text{Final volume} = \frac{800}{2} = 400 \text{ mL} = 0.4 \text{ L}$$

$$\text{Molarity} = \frac{\text{Remaining moles}}{\text{Final volume}} = \frac{0.2175}{0.4} = 0.54375 \text{ M}$$

$$\text{Molarity} = 0.54375 \text{ M} \approx 0.54 \text{ M} = 54 \times 10^{-2} \text{ M}$$

Hence, the value of $$x$$ is 54.

We have 1.80 g of solute A dissolved in 62.5 cm$$^3$$ of ethanol. The freezing point of the solution is 155.1 K, while the freezing point of pure ethanol is 156.0 K. We need to find the molar mass of solute A.

We begin by calculating the freezing point depression:

$$\Delta T_f = T_f^0 - T_f = 156.0 - 155.1 = 0.9 \text{ K}$$

Now we find the mass of ethanol (solvent). Given the volume is 62.5 cm$$^3$$ and the density is 0.80 g cm$$^{-3}$$:

$$\text{Mass of ethanol} = 62.5 \times 0.80 = 50.0 \text{ g} = 0.050 \text{ kg}$$

The freezing point depression is related to molality by:

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

where $$m$$ is the molality (moles of solute per kg of solvent) and $$K_f = 2.00$$ K kg mol$$^{-1}$$. Substituting:

$$0.9 = 2.00 \times m$$

$$m = \frac{0.9}{2.00} = 0.45 \text{ mol kg}^{-1}$$

The molality is defined as moles of solute per kg of solvent, so the number of moles of solute A is:

$$n_A = m \times \text{mass of solvent (in kg)} = 0.45 \times 0.050 = 0.0225 \text{ mol}$$

Now the molar mass of solute A is:

$$M_A = \frac{\text{mass of solute}}{n_A} = \frac{1.80}{0.0225} = 80 \text{ g mol}^{-1}$$

Hence, the correct answer is 80.

We need to find the partial pressure of hydrogen in the mixture.

The mixture contains $$40\%$$ hydrogen by mass.

Let the total mass of the mixture be $$100 \text{ g}$$.

Mass of $$H_2 = 40 \text{ g}$$

Mass of $$O_2 = 100 - 40 = 60 \text{ g}$$

Molar mass of $$H_2 = 2 \text{ g/mol}$$

Molar mass of $$O_2 = 32 \text{ g/mol}$$

$$n_{H_2} = \frac{40}{2} = 20 \text{ mol}$$

$$n_{O_2} = \frac{60}{32} = 1.875 \text{ mol}$$

Total moles = $$20 + 1.875 = 21.875 \text{ mol}$$

$$\chi_{H_2} = \frac{n_{H_2}}{n_{\text{total}}} = \frac{20}{21.875}$$

Using Dalton's law of partial pressures:

$$p_{H_2} = \chi_{H_2} \times P_{\text{total}}$$

$$= \frac{20}{21.875} \times 2.2$$

$$= \frac{20 \times 2.2}{21.875}$$

$$= \frac{44}{21.875}$$

$$= 2.0114 \approx 2 \text{ bar}$$

Hence, the partial pressure of hydrogen is 2 bar.

We need to find the percentage of dissociation of acetic acid.

The volume of acetic acid is 1.2 mL, its density is 1.02 g mL$$^{-1}$$, its molar mass is 60 g mol$$^{-1}$$, and the total volume of solution is 2.0 L. The depression in freezing point ($$\Delta T_f$$) is 0.0198 °C and K$$_f$$(H$$_2$$O) = 1.85 K kg mol$$^{-1}$$.

The mass of acetic acid is 1.2 × 1.02 = 1.224 g, and the moles of acetic acid are $$\frac{1.224}{60}$$ = 0.0204 mol.

Assuming the solution is very dilute so that 2.0 L of solution corresponds to approximately 2.0 kg of solvent, the molality (m) is $$\frac{0.0204}{2.0}$$ = 0.0102 mol kg$$^{-1}$$.

The van't Hoff factor (i) is determined from the relationship $$\Delta T_f = i \times K_f \times m$$. Substituting the known values gives $$0.0198 = i \times 1.85 \times 0.0102$$, which simplifies to $$0.0198 = i \times 0.01887$$, and hence $$i = \frac{0.0198}{0.01887} = 1.0493$$.

For acetic acid: CH$$_3$$COOH $$\rightleftharpoons$$ CH$$_3$$COO$$^-$$ + H$$^+$$, the van't Hoff factor is given by $$i = 1 + \alpha(n - 1)$$ where n = number of particles formed = 2 (one CH$$_3$$COO$$^-$$ and one H$$^+$$). Substituting gives $$1.0493 = 1 + \alpha(2 - 1)$$, and therefore $$\alpha = 0.0493$$.

The percentage dissociation is calculated as $$\alpha \times 100 = 0.0493 \times 100 \approx 5\%$$.

Therefore, the percentage of dissociation of acetic acid is 5%.

We need to find the percentage dissociation of KCl given the freezing point depression data.

Calculate molality

A 0.5% solution means 0.5 g of KCl in 100 g of solution.

Mass of solvent (water) = 100 - 0.5 = 99.5 g = 0.0995 kg

Moles of KCl = $$\frac{0.5}{74.6}$$ = 0.006702 mol

Molality (m) = $$\frac{0.006702}{0.0995}$$ = 0.06736 mol/kg

Calculate the observed van't Hoff factor (i)

Using the formula: $$\Delta T_f = i \times K_f \times m$$

$$0.24 = i \times 1.80 \times 0.06736$$

$$0.24 = i \times 0.12125$$

$$i = \frac{0.24}{0.12125} = 1.979$$

Calculate percentage dissociation

For KCl dissociating as: $$KCl \to K^+ + Cl^-$$

Number of ions (n) = 2

The van't Hoff factor is related to degree of dissociation ($$\alpha$$) by:

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

$$1.979 = 1 + \alpha(2 - 1)$$

$$1.979 = 1 + \alpha$$

$$\alpha = 0.979$$

Percentage dissociation = $$\alpha \times 100 = 97.9\% \approx 98\%$$

The correct answer is 98.

We need to find the mass of $$CO_2$$ dissolved in 1 litre of water at 298 K.

Since Henry’s Law relates the partial pressure of a gas to its mole fraction in solution, we write

$$ p_{CO_2} = K_H \cdot x_{CO_2} $$

Substituting the given values gives

$$ x_{CO_2} = \frac{p_{CO_2}}{K_H} = \frac{0.835 \text{ bar}}{1.67 \text{ kbar}} = \frac{0.835}{1670} = 5 \times 10^{-4} $$

Next, the number of moles of water in 1 litre is

$$ n_{H_2O} = \frac{1000}{18} = 55.56 \text{ mol} $$

Since $$x_{CO_2}$$ is very small, we approximate

$$ x_{CO_2} \approx \frac{n_{CO_2}}{n_{H_2O}} $$

This gives

$$ n_{CO_2} = x_{CO_2} \times n_{H_2O} = 5 \times 10^{-4} \times 55.56 = 0.02778 \text{ mol} $$

Calculating the mass of $$CO_2$$ (molar mass of $$CO_2$$ = 12 + 2(16) = 44 g/mol) yields

$$ m_{CO_2} = 0.02778 \times 44 = 1.2222 \text{ g} $$

Expressing this value in the required form results in

$$ x = 1.2222 \text{ g} \approx 1221 \times 10^{-3} \text{ g} $$

The answer is $$1221$$.

We are given elevation in boiling point for 1.5 molal glucose solution: $$\Delta T_b = 4$$ K and depression in freezing point for 4.5 molal glucose solution: $$\Delta T_f = 4$$ K.

Using the boiling point elevation formula $$\Delta T_b = K_b \times m$$ gives $$4 = K_b \times 1.5$$ and hence $$K_b = \frac{4}{1.5} = \frac{8}{3}$$ K kg mol$$^{-1}$$. Similarly, the freezing point depression formula $$\Delta T_f = K_f \times m$$ yields $$4 = K_f \times 4.5$$ so $$K_f = \frac{4}{4.5} = \frac{8}{9}$$ K kg mol$$^{-1}$$. The ratio is then $$\frac{K_b}{K_f} = \frac{8/3}{8/9} = \frac{8}{3} \times \frac{9}{8} = \frac{9}{3} = 3$$.

Therefore, the ratio $$K_b/K_f = \textbf{3}$$.

We are given that the elevation in boiling point for a 1 molal solution of non-volatile solute A is $$3\text{ K}$$ and the depression in freezing point for a 2 molal solution of A in the same solvent is $$6\text{ K}$$.

Since the elevation in boiling point is related by $$\Delta T_b = K_b \cdot m$$, substituting $$m = 1$$ molal gives $$3 = K_b \times 1$$, which yields $$K_b = 3\text{ K kg/mol}$$.

Similarly, because the depression in freezing point follows $$\Delta T_f = K_f \cdot m$$, substituting $$m = 2$$ molal leads to $$6 = K_f \times 2$$, and hence $$K_f = 3\text{ K kg/mol}$$.

Therefore, the ratio is $$\frac{K_b}{K_f} = \frac{3}{3} = 1$$.

Thus, $$K_b / K_f = 1 : 1$$, which means $$X = 1$$.

Therefore, the answer is $$\boxed{1}$$.

We need to find the normality of $$H_2SO_4$$ in the solution obtained by mixing 100 mL of 0.1 M $$H_2SO_4$$ with 50 mL of 0.1 M NaOH.

Initially, the millimoles of $$H_2SO_4$$ in 100 mL of 0.1 M $$H_2SO_4$$ is calculated as $$100 \times 0.1 = 10$$ mmol, and the millimoles of NaOH in 50 mL of 0.1 M NaOH is $$50 \times 0.1 = 5$$ mmol.

Since $$H_2SO_4$$ is a diprotic acid, it reacts with NaOH according to the equation

$$H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$$

Thus, 5 mmol of NaOH will react with $$\frac{5}{2} = 2.5$$ mmol of $$H_2SO_4$$.

After the reaction, the remaining $$H_2SO_4$$ is $$10 - 2.5 = 7.5$$ mmol.

The total volume of the mixture becomes $$100 + 50 = 150$$ mL. The remaining $$H_2SO_4$$ still has both acidic protons available, so its milliequivalents are $$7.5 \times 2 = 15$$ meq (since the n-factor of $$H_2SO_4$$ is 2).

$$\text{Normality} = \frac{15}{150} = 0.1 \text{ N} = 1 \times 10^{-1} \text{ N}$$

Hence, the answer is 1.

The mass of acetic acid (solvent) is 150 g = 0.150 kg, the mass of ascorbic acid ($$C_6H_8O_6$$, solute) is 10.2 g, the molar mass of ascorbic acid is 176 g/mol, and the cryoscopic constant $$K_f$$ of acetic acid is 3.9 K kg mol$$^{-1}$$.

Since the moles of ascorbic acid can be calculated by $$n = \frac{10.2}{176} = 0.05795 \text{ mol}$$, this represents the number of moles of solute present.

Substituting into the definition of molality gives $$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.05795}{0.150} = 0.3864 \text{ mol/kg}$$.

Next, the depression in freezing point is given by $$\Delta T_f = K_f \times m$$. Substituting the values yields $$= 3.9 \times 0.3864$$ and thus $$= 1.507 \text{ °C}$$.

Expressing this in the required form gives $$\Delta T_f = 1.507 \text{ °C} = 15.07 \times 10^{-1} \text{ °C}$$. Therefore, the value of $$x$$ is 15 (nearest integer).

The elevation in boiling point is given by:

$$\Delta T_b = K_b \times m$$

where $$K_b$$ is the ebullioscopic constant and $$m$$ is the molality.

Molality is given by:

$$m = \frac{n_{\text{solute}}}{W_{\text{solvent (kg)}}}$$

Since the same solute (2 g of non-volatile non-electrolyte) is dissolved in the same mass of solvent (200 g) for both solvents A and B:

$$m_A = m_B = m = \frac{n}{0.2}$$

The molality is the same for both solutions.

Therefore:

$$\Delta T_{b,A} = K_{b,A} \times m$$

$$\Delta T_{b,B} = K_{b,B} \times m$$

Taking the ratio:

$$\frac{\Delta T_{b,A}}{\Delta T_{b,B}} = \frac{K_{b,A}}{K_{b,B}}$$

Given that $$K_{b,A} : K_{b,B} = 1 : 8$$:

$$\frac{\Delta T_{b,A}}{\Delta T_{b,B}} = \frac{1}{8} = \frac{x}{y}$$

So $$x = 1$$ and $$y = 8$$.

Hence, the value of $$y$$ is 8.

We need to find the number of millimoles of $$O_2$$ that dissolve in 1 litre of water at 303 K using Henry's Law.

Henry's Law states: $$p = K_H \cdot x$$, where $$p$$ is the partial pressure of the gas, $$K_H$$ is the Henry's Law constant, and $$x$$ is the mole fraction of the gas in the solution.

Given: $$K_H = 46.82$$ kbar $$= 46.82 \times 10^3$$ bar, and $$p_{O_2} = 0.920$$ bar.

The mole fraction of $$O_2$$ in water is:

$$x = \frac{p}{K_H} = \frac{0.920}{46.82 \times 10^3} = \frac{0.920}{46820} = 1.9650 \times 10^{-5}$$

Now, since the solubility is very small, the mole fraction can be approximated as:

$$x \approx \frac{n_{O_2}}{n_{water}}$$

For 1 litre of water, $$n_{water} = \frac{1000}{18} = 55.556$$ mol.

$$n_{O_2} = x \times n_{water} = 1.9650 \times 10^{-5} \times 55.556 = 1.092 \times 10^{-3} \text{ mol}$$

Converting to millimoles: $$n_{O_2} = 1.092$$ mmol $$\approx 1$$ mmol (nearest integer).

Hence, the correct answer is 1.

We need to find the concentration of glucose solution (in g/L) that is isotonic with blood. For isotonic solutions, the osmotic pressure of the glucose solution must equal the osmotic pressure of blood.

The formula for osmotic pressure is:

$$ \pi = CRT $$

where $$\pi$$ is the osmotic pressure, $$C$$ is the molar concentration (mol/L), $$R$$ is the gas constant, and $$T$$ is the temperature in Kelvin.

With $$\pi = 7.47$$ bar, $$T = 300$$ K, and $$R = 0.083$$ L bar$$^{-1}$$ mol$$^{-1}$$, we solve for the molar concentration $$C$$ as follows:

$$ C = \frac{\pi}{RT} = \frac{7.47}{0.083 \times 300} $$

$$ C = \frac{7.47}{24.9} = 0.3 \text{ mol/L} $$

Converting molar concentration to g/L using the molar mass of glucose ($$M = 180$$ g/mol):

$$ \text{Concentration (g/L)} = C \times M = 0.3 \times 180 $$

$$ \text{Concentration (g/L)} = 54 $$

Therefore, the concentration of glucose solution is 54 g/L.

We need to find the value of $$x$$ given that the mole fraction of liquid B in the vapour phase is $$\frac{x}{17}$$. The vapour pressures of pure A and B are $$P_A^0 = 50$$ Torr and $$P_B^0 = 100$$ Torr respectively, and in the liquid phase the mole fractions are $$x_A = 0.3$$ and $$x_B = 0.7$$.

By Raoult's law, the partial pressures are calculated as $$P_A = x_A \times P_A^0 = 0.3 \times 50 = 15 \text{ Torr}$$ and $$P_B = x_B \times P_B^0 = 0.7 \times 100 = 70 \text{ Torr}$$. The total vapour pressure is then $$P_{total} = P_A + P_B = 15 + 70 = 85 \text{ Torr}$$.

Applying Dalton's law, the mole fraction of B in vapour is $$y_B = \frac{P_B}{P_{total}} = \frac{70}{85}$$, which simplifies to $$y_B = \frac{14}{17}$$. Since this equals $$\frac{x}{17}$$, we have $$\frac{x}{17} = \frac{14}{17}$$, giving $$x = 14$$.

We need to find the number of moles of solute A dissolved in 100 g of water such that the vapour pressure is reduced to half. Given that the mass of water (solvent) is 100 g, the vapour pressure of pure water is $$P^0 = 23.76$$ mmHg and the vapour pressure of the solution is $$P = \frac{P^0}{2} = 11.88$$ mmHg.

Using Raoult’s law, the mole fraction of the solute is given by $$\frac{P^0 - P}{P^0} = x_{\text{solute}}$$, so $$\frac{23.76 - 11.88}{23.76} = x_{\text{solute}}$$, which yields $$x_{\text{solute}} = \frac{11.88}{23.76} = 0.5$$.

The moles of water present are $$\frac{100}{18} = 5.55$$ mol. Since $$x_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{water}}}$$, we have $$0.5 = \frac{n_A}{n_A + 5.55}$$, which leads to $$0.5(n_A + 5.55) = n_A$$, then $$0.5 n_A + 2.775 = n_A$$, so $$0.5 n_A = 2.775$$ and thus $$n_A = 5.55 \text{ mol}$$.

Hence, the number of moles of solute A added is 5.55.

A solution containing $$2.5 \times 10^{-3}$$ kg of a solute dissolved in $$75 \times 10^{-3}$$ kg of water boils at $$373.535$$ K.

Mass of solute:

$$w_2 = 2.5 \times 10^{-3}\,\text{kg} = 2.5\,\text{g}$$

Mass of solvent:

$$w_1 = 75 \times 10^{-3}\,\text{kg} = 0.075\,\text{kg}$$

Boiling point elevation:

$$\Delta T_b = 373.535 - 373.15$$

$$= 0.385\,K$$

Using:

$$\Delta T_b = K_b m$$

$$m = \frac{\Delta T_b}{K_b}$$

$$m = \frac{0.385}{0.52}$$

$$= 0.7404\,\text{mol kg}^{-1}$$

Now,

$$m = \frac{w_2/M}{w_1}$$

$$0.7404 = \frac{2.5/M}{0.075}$$

$$0.7404 \times 0.075 = \frac{2.5}{M}$$

$$0.05553 = \frac{2.5}{M}$$

$$M = \frac{2.5}{0.05553}$$

$$M \approx 45\,\text{g mol}^{-1}$$

Final Answer:

$${45\,\text{g mol}^{-1}}$$

We need to calculate the osmotic pressure of a protein solution.

The mass of the protein is 2.0 g, the molar mass is $$60\,\mathrm{kg\,mol^{-1}}$$ (equivalent to $$60000\,\mathrm{g\,mol^{-1}}$$), the volume of the solution is 200 mL (0.2 L), the temperature is 27 °C (300 K), and $$R = 0.083\,\mathrm{L\,bar\,mol^{-1}\,K^{-1}}$$.

The number of moles of protein is given by

$$n = \frac{2.0}{60000} = \frac{1}{30000}\text{ mol}.$$

The osmotic pressure is determined using

$$\pi = \frac{nRT}{V}$$

Substituting the values yields

$$\pi = \frac{\frac{1}{30000} \times 0.083 \times 300}{0.2}$$

This simplifies to

$$\pi = \frac{\frac{24.9}{30000}}{0.2} = \frac{0.00083}{0.2} = 0.00415\text{ bar}.$$

Since $$1\text{ bar} = 10^5\text{ Pa}$$, converting the pressure to pascals gives

$$\pi = 0.00415 \times 10^5 = 415\text{ Pa}.$$

Therefore, the osmotic pressure is 415 Pa.

The volume of acetic acid solution is 200 mL = 0.2 L, the initial concentration is 0.2 M, the final concentration after adsorption is 0.1 M, and the mass of wood charcoal is 0.6 g.

Since volume and concentration are known, the initial moles of acetic acid are calculated as:

$$ n_{\text{initial}} = 0.2 \times 0.2 = 0.04 \text{ mol} $$

Similarly, the final moles of acetic acid after adsorption are given by:

$$ n_{\text{final}} = 0.2 \times 0.1 = 0.02 \text{ mol} $$

This gives the moles of acetic acid adsorbed as:

$$ n_{\text{adsorbed}} = 0.04 - 0.02 = 0.02 \text{ mol} $$

Substituting the molar mass of $$CH_3COOH = 60$$ g/mol gives the mass of acetic acid adsorbed:

$$ m_{\text{adsorbed}} = 0.02 \times 60 = 1.2 \text{ g} $$

From the above, the mass of acetic acid adsorbed per gram of carbon is:

$$ \frac{m_{\text{adsorbed}}}{m_{\text{carbon}}} = \frac{1.2}{0.6} = 2 \text{ g/g} $$

Therefore, the mass of acetic acid adsorbed per gram of carbon is 2 g.

We have a gaseous mixture of A and B in equilibrium with an ideal liquid solution at a total pressure of $$P = 0.8$$ atm. The mole fraction of A in the vapour phase is $$y_A = 0.5$$ and in the liquid phase is $$x_A = 0.2$$.

By Raoult's law for an ideal solution, the partial pressure of A in the vapour phase equals:

$$p_A = x_A \cdot P_A^0$$

where $$P_A^0$$ is the vapour pressure of pure liquid A.

Also, from Dalton's law of partial pressures, the partial pressure of A in the vapour is:

$$p_A = y_A \cdot P = 0.5 \times 0.8 = 0.4 \text{ atm}$$

Now using Raoult's law:

$$0.4 = 0.2 \times P_A^0$$

$$P_A^0 = \frac{0.4}{0.2} = 2 \text{ atm}$$

Hence, the correct answer is 2.

We are given that 2.5 g of a protein (composed entirely of glycine units) is dissolved in water to make 500 mL of solution, and the osmotic pressure at 300 K is $$5.03 \times 10^{-3}$$ bar.

Using the osmotic pressure formula $$\pi = cRT$$, we find $$c = \frac{\pi}{RT} = \frac{5.03 \times 10^{-3}}{0.083 \times 300} = \frac{5.03 \times 10^{-3}}{24.9} = 2.02 \times 10^{-4} \text{ mol/L}$$.

The number of moles of protein in 0.5 L is $$n = c \times V = 2.02 \times 10^{-4} \times 0.5 = 1.01 \times 10^{-4} \text{ mol}$$, and its molar mass is $$M = \frac{\text{mass}}{n} = \frac{2.5}{1.01 \times 10^{-4}} = 24752 \text{ g/mol}$$.

The molar mass of glycine ($$\text{C}_2\text{H}_5\text{NO}_2$$) is $$M_{\text{glycine}} = 2(12) + 5(1) + 14 + 2(16) = 24 + 5 + 14 + 32 = 75 \text{ g/mol}$$, and $$\text{Number of glycine units} = \frac{24752}{75} \approx 330$$.

Hence, the total number of glycine units present in the protein is 330.

Biochemical Oxygen Demand (BOD):

BOD is the amount of oxygen required by bacteria to decompose the organic matter present in a given sample of water.

Key Criteria as per NCERT:

Clean Water ($$A$$): Clean water has a low concentration of organic waste, so it requires very little oxygen for decomposition. The BOD value for clean water is less than $$5\text{ ppm}$$ ($$A < 5$$).

Highly Polluted Water ($$B$$): Highly polluted water contains a large amount of organic matter, resulting in high bacterial activity and high oxygen consumption. The BOD value for polluted water is greater than $$17\text{ ppm}$$ ($$B > 17$$).

Therefore, the expected BOD values are $$A < 5\text{ ppm}$$ and $$B > 17\text{ ppm}$$.

Hence the correct option is A.

We first recall that a lyophilic sol is one in which the dispersed phase has a definite affinity for the dispersion medium. Because of this affinity, each colloidal particle becomes surrounded by a sheath of solvent molecules. This phenomenon is called solvation (or hydration when the solvent is water).

The solvation sheath behaves like a protective layer. Since the solvent molecules are firmly attached to the surface, two adjoining particles cannot come close enough to aggregate; the intervening solvating layer keeps them apart. Mathematically, we may express the decrease in free energy accompanying solvation as $$\Delta G = \Delta H - T\Delta S < 0,$$ indicating that the process is spontaneous and favours dispersal rather than coagulation.

Because the particles remain well separated, the sol does not settle or precipitate even on standing for a long time. Hence we say that lyophilic sols are intrinsically stable. On the other hand, in a lyophobic sol the particles do not get solvated. They must rely solely on the electrostatic charge present on their surfaces for mutual repulsion. If that charge is removed (for example, by adding an electrolyte), rapid coagulation occurs. This clearly shows that solvation, and not mere charge, is the principal reason for the greater stability of lyophilic sols.

Now we examine the given options. Option A speaks of “no charge,” but we have just argued that stability arises even in the presence or absence of charge provided the particles are solvated. Option B mentions a “positive charge,” which again is not the basic cause. Option D talks of “strong electrostatic repulsion between negatively charged particles,” a feature more typical of lyophobic sols. Only Option C correctly states that the colloidal particles are solvated, and this solvation is the key to their superior stability.

Hence, the correct answer is Option C.

When $$FeCl_3$$ is added to excess of hot water, it undergoes hydrolysis to form a colloidal solution of ferric hydroxide ($$Fe(OH)_3$$):

$$FeCl_3 + 3H_2O \to Fe(OH)_3 \, (colloid) + 3HCl$$

The colloidal particles of $$Fe(OH)_3$$ preferentially adsorb $$Fe^{3+}$$ ions from the solution onto their surface (since $$Fe^{3+}$$ is a common ion with the lattice of the colloid). This adsorption of positive $$Fe^{3+}$$ ions imparts a positive charge to the colloidal particles.

Therefore, the resulting colloidal particles carry a positive charge.

The correct answer is Option (1): positive.

The Tyndall effect is the phenomenon in which a beam of light becomes visible when it passes through a medium containing particles large enough to scatter the light. The necessary condition is that the dispersed particles should have a diameter roughly between $$10^{-7}\,\text{cm}$$ and $$10^{-5}\,\text{cm}$$ so that they can scatter visible radiation. If the particles are much smaller than this range, no appreciable scattering occurs, while if they are much larger, the medium ceases to be optically homogeneous and the beam gets blocked instead of only being scattered.

Now, let us examine each of the given dispersions in the light of this particle-size criterion.

We have a true solution: in a true solution the solute particles have dimensions typically less than $$10^{-8}\,\text{cm}$$. Because these particles are far smaller than the wavelength of visible light, they do not scatter light to any observable extent. So, a true solution does not exhibit the Tyndall effect.

Next, consider a suspension: in a suspension the particles are usually larger than $$10^{-5}\,\text{cm}$$. Such coarse particles block or absorb light instead of merely scattering it, so the passage of a narrow beam cannot be seen as a luminous path. Therefore, a suspension is also unsuited for showing the Tyndall effect.

Now we look at lyophilic colloids: in a lyophilic (solvent-loving) colloid, although the particle size is in the proper colloidal range, the dispersed particles are surrounded by thick layers of the solvent. This sheath diminishes the difference between the refractive indices of the dispersed phase and the dispersion medium, and light scattering depends directly on this refractive-index contrast. Because of the reduced contrast, a lyophilic colloid shows the Tyndall effect only weakly.

Lastly, consider lyophobic colloids: in a lyophobic (solvent-hating) colloid the particles lie squarely in the colloidal size range, and they do not carry thick solvent layers. Hence the refractive-index difference between the dispersed particles and the medium remains large, giving rise to strong scattering. Consequently, the luminous path of a light beam becomes very bright and distinct; in other words, the Tyndall effect is displayed most effectively by a lyophobic colloid.

Comparing all four cases, we find that the maximum light scattering, and hence the most prominent Tyndall effect, occurs for the lyophobic colloid.

Hence, the correct answer is Option C.