Q.The percentage of helium filled in the tanks used by most scuba divers to dilute air in deep dives

Showing the 12 most recent of 21 on this concept.

1. CBSE 2026Set V11 markMCQ
   Q.The percentage of helium filled in the tanks used by most scuba divers to dilute air in deep dives

   (a) 32.1

   (b) 11.7

   (c) 74.2

   (d) 56.2

   ›Reveal solutionSolution

   The air in a deep-diving scuba tank is diluted with about 11.7% helium.

   At the high pressures experienced in deep dives, the solubility of atmospheric gases in blood increases (Henry's law, $p = K_H\,x$). Nitrogen in particular dissolves and, on rapid ascent, is released as bubbles causing the painful and dangerous condition known as bends. To minimise this, the air supplied to divers is diluted with the sparingly-soluble, chemically inert gas hel …

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2. CBSE 2026Set ANNUAL1 mark
   Q.On increasing temperature, solubility of gases in liquids ______ (fill in the blank).

   ›Reveal solutionSolution

   Dissolution of a gas in a liquid is an exothermic process, so by Le Chatelier's principle, raising the temperature shifts the equilibrium back towards the gas phase, lowering solubility.

   …

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3. CBSE 2025Set 56/4/11 markMCQ
   Q.The value of Henry's constant $K_H$ is : (A) greater for gases with higher solubility (B) greater for gases with lower solubility (C) constant for all gases (D) not related to the solubility of gases

   ›Reveal solutionSolution

   Henry’s constant $K_H$ is inversely related to gas solubility — a higher $K_H$ means lower solubility. So the correct option is (B).

   Why Henry’s Law works this way

   Henry’s Law describes the relationship between the partial pressure of a gas above a liquid and its concentration in the liquid. The law is written as:

   $p = K_H \cdot x$

   where $p$ is the partial pressure of the gas, $x$ is its mole fraction in the solution, and $K_H$ is Henry’s constant.

   The key intuition: $K_H$ is essentially a resistance to dissolution. A gas that dissolves easily (high solubility) will need only a small partial pressure to achieve a given concentration — so $K_H$ is small. Conversely, a gas that barely dissolves (low solubility) needs a large partial pressure to force even a tiny amount into solution — so $K_H$ is large.

   $p = K_H \cdot x \quad \Rightarrow \quad K_H = \frac{p}{x}$

   Step-by-step reasoning

   1. Interpret the equation For a fixed partial pressure $p$, the mole fraction $x$ of the dissolved gas is $x = p / K_H$. Since $p$ is constant, $x$ (which measures solubility) is inversely proportional to $K_H$:

   $x \propto \frac{1}{K_H}$

   2. Relate $K_H$ to solubility

      • High solubility → large $x$ → small $K_H$
      • Low solubility → small $x$ → large $K_H$

   3. Check the options

      • (A) says $K_H$ is greater for gases with higher solubility — this is the opposite of what we just found.
      • (B) says $K_H$ is greater for gases with lower solubility — this matches the inverse relationship. …

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4. CBSE 2025Set 56/5/11 markMCQ
   Q.Two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer from the codes (A), (B), (C) and (D) as given below. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true. Assertion (A) : Henry's law constant ($K_H$) decreases with increase in temperature. Reason (R) : As the temperature increases, solubility of gases in liquids decreases.

   ›Reveal solutionSolution

   Henry’s law constant $K_H$ actually increases with temperature, not decreases — so Assertion (A) is false. The Reason (R) is true: gas solubility does fall as temperature rises. The correct choice is (D).

   ---

   Why this question trips students up

   The trap here is subtle. Most of us remember that “solubility of gases decreases with temperature” — that’s drilled into us from everyday life (a cold soda fizzes more than a warm one). So Reason (R) feels solid.

   But the Assertion talks about Henry’s law constant $K_H$, not solubility directly. The two are inversely related, and that inverse relationship flips the temperature dependence. Let’s walk through it carefully.

   ---

   Step-by-step reasoning

   1. Recall Henry’s law

   Henry’s law states that at constant temperature, the partial pressure of a gas above a liquid is proportional to its mole fraction in the liquid:

   $p = K_H \cdot x$

   Here $K_H$ is the Henry’s law constant. A larger $K_H$ means that for the same partial pressure, the gas dissolves less (smaller $x$). So $K_H$ is a measure of resistance to dissolution — it’s the opposite of solubility.

   $p = K_H \cdot x \quad \Rightarrow \quad x = \frac{p}{K_H}$

   2. What happens to solubility when temperature rises?

   Dissolving a gas in a liquid is generally exothermic (heat is released). Le Chatelier’s principle tells us that raising temperature shifts equilibrium to favour the endothermic direction — which is the gas coming out of solution. So solubility $x$ decreases.

   That makes Reason (R) true.

   3. Now connect $K_H$ to temperature …

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5. CBSE 2025Set ANNUAL1 mark
   Q.State Henry's law.

   ›Reveal solutionSolution

   Henry's law states that at constant temperature, the solubility (mole fraction) of a gas in a liquid is directly proportional to the partial pressure of that gas above the liquid.

   Statement: The partial pressure of a gas in the vapour phase (p) is directly proportional to the mole fraction of the gas (x) dissolved in the solution, at a given temperature:

   $p = K_H \cdot x$

   where $K_H$ is the Henry's law constant, which is specific to the gas-solvent pair and depends on temperature.

   …

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6. CBSE 2025Set ANNUAL1 markMCQ
   Q.Dissolution of a gas in liquid is a process

   (a) with increase in enthalpy

   (b) with no change in enthalpy

   (c) with decrease in enthalpy

   (d) for which enthalpy change cannot be predicted

   ›Reveal solutionSolution

   Dissolving a gas in a liquid is an exothermic process, so it occurs with a decrease in enthalpy.

   When a gas dissolves in a liquid, gas molecules that were freely moving and widely separated come close to the solvent molecules and get surrounded by them (solvation). New, relatively strong solute–solvent (gas–liquid) interactions form as the gas molecules are trapped between solvent molecules. Energy is released when these new attractive interactions form, exactly as heat is released when a gas condenses.

   …

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7. CBSE 2025Set ANNUAL1 markMCQ
   Q.The Kₕ values of Ar(g), CO2(g), HCHO(g) and CH4(g) are 40.39, 1.67, 1.83 × 10⁻⁵ and 0.413 respectively. The correct order of their solubility is...

   (a) HCHO < CH4 < CO2 < Ar

   (b) HCHO < CO2 < CH4 < Ar

   (c) Ar < CO2 < CH4 < HCHO

   (d) Ar < CH4 < CO2 < HCHO

   ›Reveal solutionSolution

   Henry's law constant $K_H$ and gas solubility are inversely related — a HIGHER $K_H$ means LOWER solubility.

   By Henry's law, $p = K_H \cdot x$, where $x$ is the mole fraction of dissolved gas. For a fixed partial pressure $p$, a larger $K_H$ forces a smaller mole fraction $x$ to dissolve — so solubility falls as $K_H$ rises. Given: $K_H$(Ar) = 40.39, $K_H$(CO2) = 1.67, $K_H$(CH4) = 0.413, $K_H$(HCHO) = 1.83×10⁻⁵. Ranking $K_H$ from highest to lowest: A …

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8. CBSE 2024Set B1 mark
   Q.Fill in the blank: The solubility of gas in a liquid is determined by ______ law.

   ›Reveal solutionSolution

   Henry's law states that at constant temperature, the partial pressure of a gas above a solution is directly proportional to the mole fraction of the gas dissolved in the liquid.

   Henry's law: $p = K_H \cdot x$, where p is the partial pressure of the gas above the solution, x is the mole fraction of the dissolved gas in the liquid, and $K_H$ is the Henry's law constant (specific to the gas-solvent pair and temperature). A higher $K_H$ means lower solubility for a given pressure. T …

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9. CBSE 2024Set ANNUAL1 mark
   Q.The mathematical form of Henry's Law is ______.

   ›Reveal solutionSolution

   Henry's Law states that the partial pressure of a gas in the vapour phase is directly proportional to its mole fraction dissolved in the solution.

   Mathematically, Henry's Law is written as: p = KH . x …

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10. CBSE 2024Set ANNUAL1 markMCQ
    Q.The value of Henry's constant K_H :

    (a) Increases with increase in temperature

    (b) Decreases with increase in temperature

    (c) Remains constant

    (d) First increases, then decreases

    ›Reveal solutionSolution

    Gas solubility falls as temperature rises, and since $K_H$ is inversely related to solubility, $K_H$ rises with temperature.

    Henry's law states $p = K_H \cdot x$, where $p$ is the partial pressure of the gas above the solution and $x$ is its mole fraction dissolved. A larger $K_H$ means a smaller $x$ dissolves at the same pressure, i.e. lower solubility.

    …

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11. CBSE 2023Set 56/1/11 markMCQ
    Q.Low concentration of oxygen in the blood and tissues of people living at high altitude is due to : (A) high atmospheric pressure (B) low temperature (C) low atmospheric pressure (D) both low temperature and high atmospheric pressure

    ›Reveal solutionSolution

    Henry’s Law states that the solubility of a gas in a liquid is directly proportional to its partial pressure. At high altitude, low atmospheric pressure reduces the partial pressure of oxygen, lowering its concentration in blood and tissues. The correct answer is (C).

    Why Henry’s Law is the key

    The question is about oxygen concentration in blood — not about how much oxygen is in the air, but how much dissolves into the bloodstream. That’s governed by Henry’s Law, which describes the solubility of a gas in a liquid.

    $C = k_H \cdot P$

    where $C$ is the concentration of dissolved gas, $k_H$ is Henry’s constant (depends on the gas and temperature), and $P$ is the partial pressure of the gas above the liquid.

    For oxygen in blood, $k_H$ is fixed at body temperature. So the concentration of dissolved oxygen depends only on the partial pressure of oxygen in the air you breathe.

    Step-by-step reasoning

    1. At sea level, atmospheric pressure is about 1 atm. Oxygen makes up ~21% of air, so its partial pressure is $P_{O_2} \approx 0.21 \times 1 = 0.21$ atm. Henry’s Law then gives a normal dissolved oxygen concentration in blood.

    2. At high altitude (e.g., 3000–5000 m), total atmospheric pressure drops significantly — to about 0.7 atm at 3000 m, and even lower higher up. The fraction of oxygen in air stays 21%, but the partial pressure becomes $P_{O_2} \approx 0.21 \times 0.7 = 0.147$ atm.

    3. Applying Henry’s Law: Since $C = k_H \cdot P_{O_2}$, a lower $P_{O_2}$ directly gives a lower $C$. That’s why people at high altitude have low oxygen concentration in blood and tissues — not because the air has less oxygen percentage, but because the lower total pressure reduces oxygen’s partial pressure.

    4. Check the options:

       • (A) High atmospheric pressure — Wrong. High pressure would increase oxygen solubility, not decrease it. …

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12. CBSE 2023Set 56/2/11 markMCQ
    Q.Value of Henry's constant $K_H$ : (A) increases with decrease in temperature. (B) decreases with increase in temperature. (C) increases with increase in temperature. (D) remains constant.

    ›Reveal solutionSolution

    Henry’s constant $K_H$ measures gas solubility in a liquid. Since solubility decreases when temperature rises, $K_H$ must increase with temperature. The correct option is (C).

    Why Henry’s constant behaves this way

    Henry’s law states that at a constant temperature, the partial pressure of a gas above a liquid is proportional to its mole fraction in the liquid:

    $p = K_H \cdot x$

    Here $K_H$ is the Henry’s law constant. A larger $K_H$ means that for the same mole fraction $x$, the gas exerts a higher partial pressure — which implies the gas is less soluble (it prefers to stay in the vapour phase). Conversely, a smaller $K_H$ means the gas dissolves more readily.

    So $K_H$ is an inverse measure of solubility: high $K_H$ → low solubility; low $K_H$ → high solubility.

    How temperature affects solubility — and therefore $K_H$

    1. Think about dissolving a gas. When a gas dissolves in a liquid, the process is generally exothermic (heat is released). This is because gas molecules must be surrounded by solvent molecules, and the new intermolecular attractions release energy.

    2. Le Chatelier’s principle. For an exothermic dissolution, heat is a product:

    $\text{Gas} + \text{solvent} \rightleftharpoons \text{solution} + \text{heat}$

    Raising the temperature adds heat, which shifts the equilibrium backward — toward the undissolved gas. So solubility decreases as temperature rises.

    3. Connect to $K_H$. Since $K_H$ is inversely related to solubility, a decrease in solubility means $K_H$ must increase. Every textbook example confirms this: for oxygen in water, $K_H$ roughly doubles when going from 0 °C to 40 °C. …

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