Q.Why are aquatic species more comfortable in cold water in comparison to warm water?

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) …

   🔒 Start your 14-day free trial to see the full solution →
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 $T$ $$$$

   🔒 Start your 14-day free trial to see the full solution →
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. …

   🔒 Start your 14-day free trial to see the full solution →
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 …

   🔒 Start your 14-day free trial to see the full solution →
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" …

   🔒 Start your 14-day free trial to see the full solution →
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 $M_A : M_B = 5:2$. …

   🔒 Start your 14-day free trial to see the full solution →
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 $P_{\text{total}} = P_A + P_B = \frac{p}{3} + \frac{4p}{3} = \frac{5p}{3}$. …

   🔒 Start your 14-day free trial to see the full solution →
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: $$$$

   🔒 Start your 14-day free trial to see the full solution →
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. …

   🔒 Start your 14-day free trial to see the full solution →
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 $x_A = \tfrac15$, $x_B = \tfrac45$: …

    🔒 Start your 14-day free trial to see the full solution →
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. …

    🔒 Start your 14-day free trial to see the full solution →
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. …

    🔒 Start your 14-day free trial to see the full solution →