Q.At equilibrium the rate of dissolution of a solid solute in a volatile liquid solvent is __________.

Showing the 12 most recent of 33 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.The difference in bond angles between SO$_2$ and H$_2$O is (A) $12.5^\circ$ (B) $17.5^\circ$ (C) $15.0^\circ$ (D) $13.0^\circ$

   ›Reveal solutionSolution

   The bond angle in SO₂ is about $119^\circ$ (due to VSEPR: trigonal planar with one lone pair), and in H₂O it is about $104.5^\circ$ (tetrahedral with two lone pairs). The difference is $119^\circ - 104.5^\circ = 14.5^\circ$, which rounds to $15.0^\circ$, so the correct option is (C).

   The key to this question is understanding how molecular geometry and lone pairs affect bond angles. Both SO₂ and H₂O are bent molecules, but their central atoms have different numbers of lone pairs, which changes the repulsion pattern and the angle.

   In VSEPR theory, lone pairs repel more strongly than bonding pairs. The more lone pairs you have, the more the bonding pairs are squeezed together, reducing the bond angle. Let’s apply this to each molecule.

   1. SO₂ (sulfur dioxide)

      Sulfur has 6 valence electrons. In SO₂, it forms two double bonds with oxygen atoms (using 4 electrons) and retains one lone pair. That gives a steric number of 3 (two bonds + one lone pair). The electron geometry is trigonal planar, with ideal angles of $120^\circ$. But the lone pair repels the bonding pairs more than they repel each other, so the O–S–O bond angle is slightly less than $120^\circ$ — experimentally, it is $119^\circ$.

   2. H₂O (water)

      Oxygen has 6 valence electrons. It forms two single bonds with hydrogen atoms (using 2 electrons) and has two lone pairs. The steric number is 4 (two bonds + two lone pairs), so the electron geometry is tetrahedral, with ideal angles of $109.5^\circ$. The two lone pairs exert stronger repulsion, compressing the H–O–H bond angle to $104.5^\circ$.

   3. Finding the difference

      Subtract the smaller angle from the larger:

   $119^\circ - 104.5^\circ = 14.5^\circ$

   Among the given options, $14.5^\circ$ is closest to $15.0^\circ$.

   Watch out

   A common mistake is to take the ideal angles ($120^\circ$ and $109.5^\circ$) and subtract them, getting $10.5^\circ$, which is not an option. You must use the actual experimental bond angles, which account for lone-pair repulsion.

   Tip

   For quick recall: SO₂ bond angle ≈ $119^\circ$, H₂O bond angle ≈ $104.5^\circ$. The difference is always around $14-15^\circ$ in such problems.

   ✓Final answer

   The correct option is (C), with a difference of $15.0^\circ$.

2. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.2.9 g of a gas (molar mass $40\, \mathrm{g\,mol^{-1}}$) at T(K) occupied the same volume as 0.184 g of dihydrogen at $17\,^\circ\mathrm{C}$ at the same pressure. The value of T(K) is (A) 568 (B) 368 (C) 468 (D) 268

   ›Reveal solutionSolution

   Using the ideal gas law under identical pressure and volume, the number of moles of each gas must be equal; solving gives $T = 368\,\mathrm{K}$, which corresponds to option (B).

   Concept & Intuition

   The problem gives two gases at the same pressure and occupying the same volume. The ideal gas law, $PV = nRT$, tells us that when $P$ and $V$ are fixed, the product $nT$ is constant. But here the gases are different, so we must compare their moles. Since $P$ and $V$ are identical for both, the number of moles of each gas must be the same. That’s the key: equal $P$ and $V$ implies equal $n$ (because $R$ is universal). So we can set the moles of the unknown gas equal to the moles of dihydrogen ($\mathrm{H_2}$) and solve for the unknown temperature.

   Step-by-step solution

   1. Find moles of dihydrogen ($\mathrm{H_2}$) Mass of $\mathrm{H_2} = 0.184\,\mathrm{g}$. Molar mass of $\mathrm{H_2} = 2.0\,\mathrm{g\,mol^{-1}}$.

   $n_{\mathrm{H_2}} = \frac{0.184}{2.0} = 0.092\,\mathrm{mol}$

   2. Find moles of the unknown gas Mass = $2.9\,\mathrm{g}$, molar mass = $40\,\mathrm{g\,mol^{-1}}$.

   $n_{\text{gas}} = \frac{2.9}{40} = 0.0725\,\mathrm{mol}$

   3. Apply the condition of equal pressure and volume For the unknown gas at temperature $T$: $PV = n_{\text{gas}} R T$ For $\mathrm{H_2}$ at $17^\circ\mathrm{C}$ (which is $17 + 273 = 290\,\mathrm{K}$): $PV = n_{\mathrm{H_2}} R (290)$ Since $P$ and $V$ are the same, the right-hand sides are equal:

   $n_{\text{gas}} R T = n_{\mathrm{H_2}} R (290)$

   Cancel $R$:

   $n_{\text{gas}} T = n_{\mathrm{H_2}} \times 290$

   4. Solve for $T$

   $T = \frac{n_{\mathrm{H_2}} \times 290}{n_{\text{gas}}} = \frac{0.092 \times 290}{0.0725}$

   Calculate stepwise:

   $0.092 \times 290 = 26.68$

   $26.68 \div 0.0725 = 368$ (since $0.0725 \times 368 = 26.68$ exactly).

   So $T = 368\,\mathrm{K}$.

   Watch out

   A common mistake is to forget converting Celsius to Kelvin. The temperature of hydrogen is given as $17^\circ\mathrm{C}$, which is $290\,\mathrm{K}$, not $17\,\mathrm{K}$. Always use Kelvin in gas law calculations.

   Tip

   Notice that the molar masses are different, but the equality of moles is what matters here. If you mistakenly set masses equal, you’d get a wrong answer. Always go through moles when $P$ and $V$ are fixed.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

3. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.Which of the following is not correctly matched with the example mentioned in brackets? (A) Solid dispersed in gas (Smoke) (B) Solid dispersed in liquid (Paint) (C) Liquid dispersed in solid (Butter) (D) Gas dispersed in liquid (Cloud)

   ›Reveal solutionSolution

   The question asks which colloid type is mismatched with its example. The key is to classify each example by the physical state of the dispersed phase and the dispersion medium. The mismatched pair is Cloud — it is a liquid-in-gas colloid, not gas-in-liquid. The correct option is (D).

   ---

   The concept here is colloidal classification by phase. A colloid is a mixture where one substance (the dispersed phase) is finely distributed throughout another (the dispersion medium). The trick is to identify the state of matter of each part — solid, liquid, or gas — and then match it to the given example.

   Let’s check each option step by step:

   1. Option (A): Solid dispersed in gas (Smoke)

      • Smoke consists of tiny solid particles (e.g., carbon or ash) suspended in air (a gas).
      • Dispersed phase: solid; dispersion medium: gas.
      • This matches perfectly. ✓

   2. Option (B): Solid dispersed in liquid (Paint)

      • Paint is a mixture of solid pigment particles dispersed in a liquid binder (like oil or water).
      • Dispersed phase: solid; dispersion medium: liquid.
      • Correct match. ✓

   3. Option (C): Liquid dispersed in solid (Butter)

      • Butter is a water-in-oil emulsion where tiny droplets of water (liquid) are trapped in a solid fat matrix.
      • Dispersed phase: liquid; dispersion medium: solid.
      • This is correct. ✓

   4. Option (D): Gas dispersed in liquid (Cloud)

      • A cloud is made of tiny water droplets (liquid) suspended in air (gas).
      • So the dispersed phase is liquid, and the dispersion medium is gas.
      • The given description says “gas dispersed in liquid” — that would be something like foam (e.g., whipped cream or shaving cream).
      • Therefore, this is incorrectly matched. ✗

   Watch out

   A common mistake is to think of clouds as “gas in liquid” because they look like floating gas. But clouds are actually liquid droplets (or ice crystals) in air — a liquid-in-gas colloid (aerosol). The classic gas-in-liquid colloid is a foam, like soap bubbles.

   Tip

   To avoid confusion, remember: the first word in the description is the dispersed phase, the second is the medium. For “Gas dispersed in liquid,” the gas is inside the liquid — think of fizzy soda or foam.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

4. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.The following graph is obtained by taking partial pressure of HCl

   (g) on y-axis and mole fraction of HCl

   (g) in cyclohexane on x-axis. The slope (m) of the graph is [FIGURE] (A) Henry constant ($K_H$) of cyclohexane (B) $K_H$ of HCl

   (g) (C) Square root of $K_H$ of HCl

   (g) (D) $\dfrac{1}{K_H(\text{HCl})}$

   ›Reveal solutionSolution

   The graph plots partial pressure of HCl (y-axis) vs. mole fraction of HCl in cyclohexane (x-axis). According to Henry’s law, the slope equals Henry’s constant for HCl in cyclohexane, so the correct option is (B).

   ---

   The key here is Henry’s law, which describes the solubility of a gas in a liquid. For a dilute solution, the partial pressure of the gas above the liquid is directly proportional to its mole fraction in the liquid:

   $P_{\text{HCl}} = K_H \cdot x_{\text{HCl}}$

   where $K_H$ is the Henry’s law constant. The constant is specific to the gas–solvent pair. In this graph, the y-axis is $P_{\text{HCl}}$ and the x-axis is $x_{\text{HCl}}$, so the slope $m$ is exactly $K_H$.

   Now, which substance does this $K_H$ belong to? The graph is for HCl gas dissolved in cyclohexane. The constant describes how much HCl partitions between the gas phase and the liquid phase. That is the Henry’s constant of HCl in cyclohexane — not of cyclohexane itself.

   Let’s walk through the reasoning step by step.

   1. Identify the axes and the law The y-axis is partial pressure of HCl gas, $P_{\text{HCl}}$. The x-axis is mole fraction of HCl in the liquid, $x_{\text{HCl}}$. Henry’s law states:

   $P_{\text{HCl}} = K_H \cdot x_{\text{HCl}}$

   This is a linear relation through the origin (for dilute solutions). The slope is $K_H$.

   2. Interpret the slope

      The slope $m = \frac{\Delta P_{\text{HCl}}}{\Delta x_{\text{HCl}}}$. Since the graph is a straight line through the origin, the slope is constant and equals the proportionality constant in Henry’s law.

   3. Assign the constant to the correct species

      Henry’s constant is always defined for the solute (the gas being dissolved), not the solvent. Here, HCl is the solute; cyclohexane is the solvent. So $K_H$ is the Henry’s constant of HCl in cyclohexane.

   4. Match with the options

      • (A) says “Henry constant of cyclohexane” — incorrect, because cyclohexane is the solvent.
      • (B) says “$K_H$ of HCl (g)” — correct.
      • (C) says “square root of $K_H$ of HCl” — no square root appears in Henry’s law.
      • (D) says “$1/K_H(\text{HCl})$” — that would be the reciprocal, not the slope.

   Watch out

   A common mistake is to think the slope is the reciprocal of Henry’s constant, or to confuse which substance the constant refers to. Remember: the graph’s slope is directly $K_H$ for the gas whose pressure is plotted.

   Tip

   If the axes were swapped (mole fraction on y, pressure on x), the slope would be $1/K_H$. Always check which variable is on which axis.

   ---

   ✓Final answer

   The correct option is (B).

   ANSWER: B

5. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.Identify the correct statements from the following I. Hydrogen bonding is present in liquid water and solid water (ice) II. Hydrogenation of vegetable oils give a fat called vanaspati III. Water present in BaCl$_2$.2H$_2$O belongs to the type "interstitial water" (A) I, III only (B) I, II, III (C) I, II only (D) II, III only

   ›Reveal solutionSolution

   All three statements are correct: hydrogen bonding is present in both liquid and solid water, hydrogenation of vegetable oils produces vanaspati, and the water in BaCl$_2$.2H$_2$O is interstitial. The correct option is (B).

   Let's analyze each statement to determine its correctness.

   Concept and Intuition

   This question tests your understanding of fundamental concepts in chemistry:

   • Hydrogen Bonding: A strong intermolecular force crucial for the properties of water.
   • Hydrogenation: A chemical reaction used to modify the properties of fats and oils.
   • Water of Crystallization: The different ways water molecules can be incorporated into crystal structures.

   We will evaluate each statement based on these principles.

   Step-by-Step Evaluation

   1. Statement I: Hydrogen bonding is present in liquid water and solid water (ice)

   • Reasoning: Hydrogen bonding is a special type of dipole-dipole interaction that occurs when a hydrogen atom covalently bonded to a highly electronegative atom (like oxygen, nitrogen, or fluorine) is attracted to another highly electronegative atom in a different molecule.
   • In water ($H_2O$), oxygen is highly electronegative and forms covalent bonds with two hydrogen atoms. This creates a strong partial positive charge on the hydrogen atoms and a strong partial negative charge on the oxygen atom.
   • Liquid Water: In liquid water, molecules are constantly moving, but they form a dynamic network of hydrogen bonds. Each water molecule can form up to four hydrogen bonds with neighboring water molecules (two through its hydrogen atoms and two through the lone pairs on its oxygen atom). These bonds are continuously breaking and reforming.
   • Solid Water (Ice): In ice, water molecules arrange themselves into a highly ordered, open, hexagonal crystalline structure. Each water molecule is tetrahedrally hydrogen-bonded to four other water molecules. This rigid, extensive network of hydrogen bonds is responsible for ice being less dense than liquid water.
   • Conclusion: Statement I is correct.

   2. Statement II: Hydrogenation of vegetable oils give a fat called vanaspati

   • Reasoning: Vegetable oils are typically unsaturated fats, meaning their fatty acid chains contain one or more carbon-carbon double bonds ($C=C$). These double bonds give them a bent structure, preventing close packing and making them liquid at room temperature.
   • Hydrogenation is a chemical process where hydrogen gas ($H_2$) is added across these carbon-carbon double bonds in the presence of a catalyst (commonly nickel, palladium, or platinum). This converts the unsaturated fatty acids into saturated fatty acids (containing only $C-C$ single bonds).
   • Product: The resulting saturated fats have straighter chains, allowing them to pack more closely. This changes their physical state from liquid oil to a solid or semi-solid fat at room temperature. This product is commonly known as vanaspati ghee (or hydrogenated vegetable oil), which is used as a cooking medium and in the production of margarine.
   • Conclusion: Statement II is correct.

   3. Statement III: Water present in BaCl$_2$.2H$_2$O belongs to the type "interstitial water"

   • Reasoning: Water of crystallization refers to water molecules that are an integral part of the crystal structure of a compound. There are different types based on how the water molecules are incorporated:
     • Coordinated water: Water molecules directly bonded to the metal ion in the coordination sphere (e.g., in $[Cu(H_2O)_4]SO_4 \cdot H_2O$, four water molecules are coordinated to copper).
     • Lattice or Interstitial water: Water molecules that occupy specific positions within the crystal lattice, filling voids or interstitial spaces. They are not directly bonded to the metal ion but are held by hydrogen bonds to other water molecules, anions, or cations in the lattice.
     • Hydrogen-bonded water: Water molecules held by hydrogen bonds to other water molecules or to anions/cations, but not necessarily occupying specific interstitial sites or being coordinated to a metal ion. This category often overlaps with interstitial water.
   • In BaCl$_2$.2H$_2$O (barium chloride dihydrate), the water molecules are not directly coordinated to the Ba$^{2+}$ ion. Instead, they are located in the spaces (interstices) within the crystal lattice and are held there by hydrogen bonds to the chloride ions and other water molecules. This arrangement is characteristic of interstitial water.
   • Conclusion: Statement III is correct.

   Since all three statements (I, II, and III) are correct, the option that includes all of them is the correct answer.

   ✓Final answer

   All three statements are correct, so the correct option is (B).

6. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.At T(K), the pressure of two ideal gases A and B is in 2:5 ratio ($p_A : p_B = 2:5$). At this temperature, their density is same. Molar mass ratio ($M_A : M_B$) is (A) $5:2$ (B) $1:2$ (C) $5:1$ (D) $25:4$

   ›Reveal solutionSolution

   Using the ideal gas law in density form $pM = \rho RT$, with equal densities and given pressure ratio, the molar mass ratio is the inverse of the pressure ratio, giving $M_A : M_B = 5:2$.

   The key concept is the density form of the ideal gas law. For an ideal gas, $pV = nRT$. Since $n = \frac{m}{M}$ (mass over molar mass), we can write $pV = \frac{m}{M}RT$. Rearranging gives $pM = \frac{m}{V}RT = \rho RT$, where $\rho$ is density. This form directly links pressure, molar mass, and density at a fixed temperature — perfect for comparing two gases when density is the same.

   1. Write the relation for each gas.

      For gas A: $p_A M_A = \rho_A R T$

      For gas B: $p_B M_B = \rho_B R T$

      Since the temperature $T$ and the gas constant $R$ are identical, and the problem states $\rho_A = \rho_B$, we can set the right-hand sides equal.

   2. Equate the two expressions.

      Because $\rho_A R T = \rho_B R T$, we have:

   $p_A M_A = p_B M_B$

   3. Use the given pressure ratio. The ratio $p_A : p_B = 2:5$ means $\frac{p_A}{p_B} = \frac{2}{5}$. From $p_A M_A = p_B M_B$, we get:

   $\frac{M_A}{M_B} = \frac{p_B}{p_A} = \frac{5}{2}$

   So $M_A : M_B = 5:2$.

   Watch out

   A common mistake is to think that higher pressure means higher molar mass when density is equal. Actually, from $pM = \rho RT$, pressure and molar mass are inversely proportional at constant density and temperature. So the gas with lower pressure has the higher molar mass.

   Tip

   You can also think of it as: if two gases have the same density at the same temperature, the one with lower pressure must have heavier molecules to pack the same mass into the same volume.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

7. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If two liquids A and B have $P_A^\circ : P_B^\circ = 1 : 2$ and have mole fraction in solution as $1 : 2$, then mole fraction of B in vapour phase is (A) 0.2 (B) 0.8 (C) 0.4 (D) 0.6

   ›Reveal solutionSolution

   Using Raoult’s law and Dalton’s law, the mole fraction of B in the vapour phase is found to be 0.8, so the correct option is (B).

   Concept & Intuition

   This problem is about the relationship between the composition of a liquid mixture and the composition of the vapour above it. The key idea: in an ideal solution, the partial pressure of each component in the vapour is given by Raoult’s law ($P_i = x_i P_i^\circ$), and the total pressure is the sum of these partial pressures. The mole fraction of a component in the vapour phase is then the ratio of its partial pressure to the total pressure. So we just need to compute these pressures from the given ratios.

   Step-by-step solution

   1. Assign variables from the given ratios

      The ratio of pure vapour pressures is $P_A^\circ : P_B^\circ = 1 : 2$. Let $P_A^\circ = p$ and $P_B^\circ = 2p$ for some $p$.

      The mole fraction ratio in the liquid is $x_A : x_B = 1 : 2$. Since $x_A + x_B = 1$, we have $x_A = \frac{1}{3}$ and $x_B = \frac{2}{3}$.

   2. Apply Raoult’s law to find partial pressures

      Raoult’s law: $P_A = x_A P_A^\circ$ and $P_B = x_B P_B^\circ$.

      So

   $P_A = \frac{1}{3} \cdot p = \frac{p}{3}, \qquad P_B = \frac{2}{3} \cdot 2p = \frac{4p}{3}.$

   3. Find total vapour pressure

      Total pressure $P_{\text{total}} = P_A + P_B = \frac{p}{3} + \frac{4p}{3} = \frac{5p}{3}$.

   4. Compute mole fraction of B in vapour

      By Dalton’s law, the mole fraction of B in the vapour phase is

   $y_B = \frac{P_B}{P_{\text{total}}} = \frac{4p/3}{5p/3} = \frac{4}{5} = 0.8.$

   Tip

   Notice that the $p$ cancels out — you never need the actual vapour pressures, only their ratio. This is a common shortcut in such problems.

   Watch out

   A common mistake is to confuse the liquid mole fraction $x_B$ with the vapour mole fraction $y_B$. Here $x_B = 2/3 \approx 0.667$, but $y_B = 0.8$ — the vapour is richer in the more volatile component (B has higher $P^\circ$).

   ✓Final answer

   The correct option is (B).

   ANSWER: B

8. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.A substance has a density of $2\ \mathrm{g\ cm^{-3}}$. It crystallizes in the fcc crystal with an edge length of $600\ \mathrm{pm}$. The molar mass of the substance (in $\mathrm{g\ mol^{-1}}$) is ($N_A = 6 \times 10^{23}\ \mathrm{mol^{-1}}$) (A) 54.8 (B) 64.8 (C) 74.8 (D) 84.7

   ›Reveal solutionSolution

   Using the density formula for a crystal, we relate density, number of atoms per unit cell, molar mass, and edge length. For fcc, $Z = 4$; solving gives molar mass ≈ 64.8 g mol⁻¹, matching option (B).

   The key idea is that the density of a crystalline solid is given by

   $\rho = \frac{Z \cdot M}{N_A \cdot a^3}$

   where $Z$ is the number of atoms per unit cell, $M$ is the molar mass, $N_A$ is Avogadro’s number, and $a$ is the edge length. For an fcc lattice, $Z = 4$. We are given $\rho = 2\ \mathrm{g\,cm^{-3}}$, $a = 600\ \mathrm{pm}$, and $N_A = 6 \times 10^{23}\ \mathrm{mol^{-1}}$. We solve for $M$.

   1. Convert edge length to cm The edge length is $600\ \mathrm{pm}$. Since $1\ \mathrm{pm} = 10^{-12}\ \mathrm{m}$ and $1\ \mathrm{m} = 100\ \mathrm{cm}$, we have

   $a = 600 \times 10^{-12}\ \mathrm{m} = 600 \times 10^{-10}\ \mathrm{cm} = 6 \times 10^{-8}\ \mathrm{cm}.$

   2. Write the density formula

   $\rho = \frac{Z \cdot M}{N_A \cdot a^3}$

   Rearranging for $M$:

   $M = \frac{\rho \cdot N_A \cdot a^3}{Z}.$

   3. Substitute the known values

      • $\rho = 2\ \mathrm{g\,cm^{-3}}$
      • $N_A = 6 \times 10^{23}\ \mathrm{mol^{-1}}$
      • $a = 6 \times 10^{-8}\ \mathrm{cm}$ → $a^3 = (6 \times 10^{-8})^3 = 216 \times 10^{-24} = 2.16 \times 10^{-22}\ \mathrm{cm^3}$
      • $Z = 4$ (fcc)

      So

   $M = \frac{2 \times (6 \times 10^{23}) \times (2.16 \times 10^{-22})}{4}.$

   4. Simplify step by step First, multiply the numerator:

   $2 \times 6 \times 10^{23} = 12 \times 10^{23}$

   Then multiply by $2.16 \times 10^{-22}$:

   $12 \times 10^{23} \times 2.16 \times 10^{-22} = 12 \times 2.16 \times 10^{1} = 25.92 \times 10 = 259.2$

   Now divide by 4:

   $M = \frac{259.2}{4} = 64.8\ \mathrm{g\,mol^{-1}}.$

   Tip

   A common mistake is forgetting to convert pm to cm. Since density is in g cm⁻³, the edge length must be in cm. Also, note that $1\ \mathrm{pm} = 10^{-10}\ \mathrm{cm}$, so $600\ \mathrm{pm} = 6 \times 10^{-8}\ \mathrm{cm}$.

   Watch out

   Some students use $Z = 1$ (simple cubic) or $Z = 2$ (bcc) by accident. For fcc, always remember $Z = 4$.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

9. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.Diborane on hydrolysis gives a compound X. The correct statements about X are I. It is a tribasic acid II. It is a weak monobasic acid III. It has a layer structure IV. It is highly soluble in water (A) I & III (B) II & III (C) II & IV (D) I & IV

   ›Reveal solutionSolution

   Diborane hydrolysis yields boric acid (H₃BO₃), which is a weak monobasic acid (not tribasic) and has a layered crystal structure; thus the correct statements are II and III, corresponding to option (B).

   Concept & Intuition

   Diborane (B₂H₆) reacts violently with water to give boric acid (H₃BO₃) and hydrogen gas. The key is understanding the actual acidic behavior of boric acid: despite having three hydroxyl groups, it does not donate three protons. Instead, it acts as a Lewis acid, accepting an OH⁻ from water to form [B(OH)₄]⁻, releasing only one H⁺. That makes it a weak monobasic acid, not a tribasic one. Also, boric acid crystallizes in a layered structure held together by hydrogen bonds, which explains its slippery feel and certain physical properties.

   1. Hydrolysis reaction Diborane reacts with water:

   $\text{B}_2\text{H}_6 + 6\text{H}_2\text{O} \rightarrow 2\text{H}_3\text{BO}_3 + 6\text{H}_2$

   So compound X is boric acid, $\text{H}_3\text{BO}_3$.

   2. Acidic nature – why it’s monobasic, not tribasic Boric acid does not ionize by losing H⁺ from its O–H bonds. Instead, it accepts a hydroxide ion from water:

   $\text{B(OH)}_3 + \text{H}_2\text{O} \rightleftharpoons [\text{B(OH)}_4]^- + \text{H}^+$

   Only one H⁺ is produced per molecule. Hence it is a weak monobasic acid (statement II is true, statement I is false).

   Watch out

   A common mistake is to count the three OH groups and assume three acidic protons. But boron’s electron deficiency makes it a Lewis acid, not a Brønsted acid.

   3. Structure – layered crystal

      In the solid state, boric acid molecules form hydrogen-bonded sheets (layers) with a hexagonal arrangement. These layers stack on top of each other with weak van der Waals forces. This layer structure gives boric acid its characteristic flaky, slippery feel (statement III is true).

   4. Solubility in water

      Boric acid is moderately soluble in water (about 5.7 g/100 mL at 25°C). It is not “highly soluble” like, say, sugar or NaCl. Statement IV (“highly soluble”) is therefore false.

   5. Matching the options

      • True statements: II (weak monobasic acid) and III (layer structure).
      • This corresponds to option (B).

   ✓Final answer

   The correct option is (B).

   ANSWER: B

10. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
    Q.Two liquids ‘A’ and ‘B’ form an ideal solution. At 300 K, the vapour pressure of a solution containing 1 mole of ‘A’ and 3 moles of ‘B’ is 550 mm Hg. At the same temperature, if one more mole of ‘B’ is added to the solution, the vapour pressure of solution increases to 560 mm Hg. Then the ratio of vapour pressures of A and B in their pure state is (A) 1 : 3 (B) 3 : 1 (C) 2 : 3 (D) 3 : 2

    ›Reveal solutionSolution

    Solving the two Raoult's-law equations gives $P_A^0 = 400$ and $P_B^0 = 600~\text{mm Hg}$, so $P_A^0 : P_B^0 = 2 : 3$ — option (C).

    Concept

    For an ideal solution, the total vapour pressure is the mole-fraction-weighted sum of the pure vapour pressures:

    $P = x_A P_A^0 + x_B P_B^0$

    Two different compositions give two equations in the two unknowns.

    Solution

    First solution — 1 mol A, 3 mol B, so $x_A = \tfrac14$, $x_B = \tfrac34$:

    $\tfrac14 P_A^0 + \tfrac34 P_B^0 = 550 \;\Rightarrow\; P_A^0 + 3P_B^0 = 2200 \quad (1)$

    Second solution — add 1 mol B: 1 mol A, 4 mol B, so $x_A = \tfrac15$, $x_B = \tfrac45$:

    $\tfrac15 P_A^0 + \tfrac45 P_B^0 = 560 \;\Rightarrow\; P_A^0 + 4P_B^0 = 2800 \quad (2)$

    Subtract (1) from (2):

    $P_B^0 = 600~\text{mm Hg}$

    Back-substitute into (1):

    $P_A^0 = 2200 - 3(600) = 400~\text{mm Hg}$

    Ratio:

    $P_A^0 : P_B^0 = 400 : 600 = 2 : 3$

    ✓Final answer

    (C) 2 : 3.

11. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
    Q.Noble gas ‘X’ is used as a diluent for oxygen in modern diving apparatus and noble gas ‘Y’ is used mainly to provide an inert atmosphere in high temperature metallurgical processes. ‘Y’ and ‘X’ are respectively? (A) He, Ar (B) Ar, He (C) He, Kr (D) Ar, Kr

    ›Reveal solutionSolution

    The key is matching each noble gas to its specific industrial use: helium (He) is the diluent for oxygen in diving, and argon (Ar) provides the inert atmosphere in high‑temperature metallurgy. Thus the pair is Ar, He — option (B).

    Concept & Intuition

    Noble gases are prized for their chemical inertness, but each has unique physical properties that suit particular applications.

    • Helium (He) is extremely light (low density) and has very low solubility in blood. This makes it ideal as a diluent for oxygen in deep‑sea diving, replacing nitrogen to avoid decompression sickness (“the bends”) and reducing breathing resistance.
    • Argon (Ar) is denser than air, cheap, and readily available. It is the most common inert gas used in welding and high‑temperature metallurgy (e.g., arc welding of aluminium, titanium processing) because it forms a protective blanket that prevents oxidation without reacting with molten metals.

    Now let’s confirm step by step.

    1. Identify noble gas ‘X’ (diluent for oxygen in diving)

       • In standard compressed air, nitrogen dissolves under pressure and can cause narcosis and decompression sickness.
       • Helium is used to replace nitrogen because it is much less soluble in blood and diffuses quickly, reducing these risks.
       • Neon and krypton are too expensive or have undesirable properties; argon is denser than air and would increase breathing resistance.
       • Therefore, X = He.

    2. Identify noble gas ‘Y’ (inert atmosphere in high‑temperature metallurgy)

       • Argon is the most widely used shielding gas in arc welding, casting, and heat treatment of reactive metals (titanium, zirconium).
       • Helium is also used in welding but is less common for metallurgy because it is lighter and escapes more easily; argon’s higher density provides a better blanket.
       • Krypton is too rare and costly for bulk industrial use.
       • Therefore, Y = Ar.

    3. Match the order in the question

       • The question asks: “Y and X are respectively?”
       • So Y = Ar, X = He → the pair is Ar, He.

    4. Check the options

       • (A) He, Ar → wrong order.
       • (B) Ar, He → correct.
       • (C) He, Kr → both wrong.
       • (D) Ar, Kr → Y correct, X wrong.

    Watch out

    A common mistake is to reverse the order: the question explicitly says “Y and X are respectively”, so the first answer is Y, the second is X. Option (A) gives He, Ar — that would be X, Y, not Y, X.

    Tip

    Remember: Helium for diving (light, low solubility), Argon for welding (dense, cheap, inert). This pairing is a classic fact in chemistry of noble gases.

    ✓Final answer

    The correct option is (B).

    ANSWER: B

12. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
    Q.Which of the following is a lung irritant that can lead to an acute respiratory disease in children? (A) CO (B) SO$_2$ (C) CO$_2$ (D) NO$_2$

    ›Reveal solutionSolution

    NO$_2$ is the textbook lung irritant that can cause acute respiratory disease in children — option (D).

    Concept. Oxides of nitrogen, especially NO$_2$, are key air pollutants formed in high-temperature combustion and in photochemical smog. NO$_2$ is a reddish-brown, highly reactive gas that attacks respiratory tissue.

    Reasoning through the options.

    • CO (A): toxic because it binds haemoglobin to form carboxyhaemoglobin, impairing oxygen transport — a blood-level poison, not primarily a lung irritant.
    • SO$_2$ (B): irritates eyes and the respiratory tract and aggravates asthma, but the specific textbook statement about acute respiratory disease in children is attached to NO$_2$.
    • CO$_2$ (C): not a lung irritant at ambient levels; its concern is the greenhouse effect.
    • NO$_2$ (D): per the standard (NCERT) environmental chemistry text, elevated NO$_2$ damages the lungs — it "is a lung irritant that can lead to an acute respiratory disease in children." It also injures plant leaves and retards photosynthesis.

    ✓Final answer

    The lung irritant that can lead to an acute respiratory disease in children is NO$_2$ — the correct option is (D).

    ANSWER: D