Q.$H_2S$, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of $H_2S$ in water at STP is 0.195 m, calculate Henry's law constant.

Showing the 12 most recent of 16 on this concept.

1. KCET 2025Set D-41 markMCQ
   Q.Variation of solubility with temperature T for a gas in liquid is shown by the following graphs. The correct representation is: (A) (B) (C) (D)

   ›Reveal solutionSolution

   Henry's law makes gas solubility fall as temperature rises — a straight line sloping DOWNWARD, not up, flat, or a hump.

   Why solubility decreases with temperature for a gas. Dissolving a gas in a liquid is generally exothermic. As temperature rises, the dissolved gas molecules gain kinetic energy and increasingly escape back into the gas phase, so less gas stays dissolved at equilibrium — the everyday example is a fizzy drink going flat faster when warm.

   Ruling out the other shapes: an upward line (A) or hump (C) would mean solubility rises with heating, which is the opposite of the real gas-solubility trend; a flat line (B) would mean no temperature dependence at all, which contradicts the well-established Henry's-law relationship.

   ✓Final answer

   Option (D) — the downward-sloping line.

2. KCET 2025Set D-41 markMCQ
   Q.If $\mathrm{N_2}$ gas is bubbled through water at 293 K, how many moles of $\mathrm{N_2}$ gas would dissolve in 1 litre of water? Assume that $\mathrm{N_2}$ exerts a partial pressure of 0.987 bar. [Given $K_H$ for $\mathrm{N_2}$ at 293 K is $76.48$ K bar] (A) $0.716 \times 10^{-3}$ (B) $7.16 \times 10^{-5}$ (C) $7.16 \times 10^{-4}$ (D) $7.16 \times 10^{-3}$

   ›Reveal solutionSolution

   Use Henry's law to get the mole fraction of dissolved $\mathrm{N_2}$, then convert that mole fraction to moles using the $\approx 55.5$ mol of water in 1 litre.

   Step 1 — Henry's law.

   The solubility of a gas in a liquid at a given temperature is proportional to its partial pressure above the liquid:

   $p = K_H \cdot x$

   where $x$ is the mole fraction of the gas in solution and $K_H$ is the Henry's-law constant. Rearranged:

   $x = \frac{p}{K_H}$

   Step 2 — Substitute (watch the units on $K_H$).

   The constant is given as $76.48\ \mathrm{K\,bar}$, i.e. $76.48$ kilobar $= 76.48\times10^3\ \mathrm{bar} = 76{,}480\ \mathrm{bar}$ (this is the standard NCERT value for $\mathrm{N_2}$ at 293 K).

   $x_{\mathrm{N_2}} = \frac{0.987\ \mathrm{bar}}{76{,}480\ \mathrm{bar}} = 1.29\times10^{-5}$

   A very small number — nitrogen is only sparingly soluble in water, as expected.

   Step 3 — Moles of water in 1 litre.

   $n_{\mathrm{H_2O}} = \frac{1000\ \mathrm{g}}{18\ \mathrm{g\,mol^{-1}}} = 55.5\ \mathrm{mol}$

   Step 4 — Convert mole fraction to moles of $\mathrm{N_2}$.

   By definition

   $x_{\mathrm{N_2}} = \frac{n_{\mathrm{N_2}}}{n_{\mathrm{N_2}} + n_{\mathrm{H_2O}}} \approx \frac{n_{\mathrm{N_2}}}{n_{\mathrm{H_2O}}}$

   The approximation is excellent because $n_{\mathrm{N_2}} \lll n_{\mathrm{H_2O}}$ (we are about to find $n_{\mathrm{N_2}} \sim 10^{-4}$ against $55.5$). Hence

   $n_{\mathrm{N_2}} = x_{\mathrm{N_2}} \times n_{\mathrm{H_2O}} = (1.29\times10^{-5})(55.5)$

   $n_{\mathrm{N_2}} = 7.16\times10^{-4}\ \mathrm{mol}$

   Step 5 — Check the options.

   $7.16\times10^{-4}$ is option (C). Option (A), $0.716\times10^{-3}$, is numerically the same number ($0.716\times10^{-3} = 7.16\times10^{-4}$) — but the standard NCERT-printed form of this answer, and the form matching the option set's convention, is (C). Options (B) and (D) are out by a factor of ten either way, which is what you get by mishandling the $\mathrm{k\,bar}$ in $K_H$ or forgetting the 55.5.

   (Physical note: the everyday consequence of this small solubility is decompression sickness — divers breathing high-pressure air dissolve extra $\mathrm{N_2}$ in their blood, and it bubbles out if they surface too fast.)

   ✓Final answer

   The correct option is (C) — $7.16 \times 10^{-4}$ mol.

   ANSWER: C

3. COMEDK 2025Set 2025-E1 markMCQ
   Q.The ratio of $\mathrm{N}_2$ and $\mathrm{O}_2$ gases in the atmosphere is $4: 1$. The ratio of the mole fractions of the dissolved gases $\mathrm{N}_2$ : $\mathrm{O}_2$ in rain water will be approximately ........ (At $293 \mathrm{~K}, \mathrm{~K}_{\mathrm{H}}$ for Nitrogen and Oxygen in kbar units are 76.48 and 34.86 respectively.) (A) $3: 1$ (B) $2: 1$ (C) $4: 1$ (D) $1: 4$

   ›Reveal solutionSolution

   Henry’s law says the mole fraction of a dissolved gas is proportional to its partial pressure times its Henry’s constant. Using the given atmospheric ratio and Henry’s constants, the dissolved N₂:O₂ ratio comes out to about 2:1, so option (B) is correct.

   Concept & Intuition

   Rainwater is in contact with air, so gases dissolve according to Henry’s law:

   $x_{\text{gas}} = \frac{p_{\text{gas}}}{K_H}$

   where $x_{\text{gas}}$ is the mole fraction in the liquid, $p_{\text{gas}}$ is the partial pressure in the gas phase, and $K_H$ is Henry’s constant (here given in kbar). The atmospheric ratio N₂:O₂ is 4:1 by volume, which means the partial pressures are in the same ratio (since total pressure ≈ 1 atm, but we only need the ratio). The dissolved ratio is not simply 4:1 because O₂ dissolves more readily (lower $K_H$). We must compute the ratio of $p/K_H$ for each gas.

   Step-by-step

   1. Set up partial pressures In dry air, N₂ and O₂ are in volume ratio 4:1. Since partial pressure is proportional to mole fraction in the gas,

   $p_{\mathrm{N_2}} : p_{\mathrm{O_2}} = 4 : 1$

   Let $p_{\mathrm{O_2}} = P$, then $p_{\mathrm{N_2}} = 4P$.

   2. Apply Henry’s law for each gas Henry’s law:

   $x_{\mathrm{N_2}} = \frac{p_{\mathrm{N_2}}}{K_{H,\mathrm{N_2}}}, \quad x_{\mathrm{O_2}} = \frac{p_{\mathrm{O_2}}}{K_{H,\mathrm{O_2}}}$

   Given $K_{H,\mathrm{N_2}} = 76.48$ kbar and $K_{H,\mathrm{O_2}} = 34.86$ kbar.

   3. Find the ratio of dissolved mole fractions

   $\frac{x_{\mathrm{N_2}}}{x_{\mathrm{O_2}}} = \frac{p_{\mathrm{N_2}} / K_{H,\mathrm{N_2}}}{p_{\mathrm{O_2}} / K_{H,\mathrm{O_2}}} = \frac{p_{\mathrm{N_2}}}{p_{\mathrm{O_2}}} \cdot \frac{K_{H,\mathrm{O_2}}}{K_{H,\mathrm{N_2}}}$

   Substitute the partial pressure ratio $4/1$:

   $\frac{x_{\mathrm{N_2}}}{x_{\mathrm{O_2}}} = 4 \times \frac{34.86}{76.48}$

   4. Calculate numerically

   $\frac{34.86}{76.48} \approx 0.4558$

   Then

   $4 \times 0.4558 \approx 1.823$

   So the ratio is roughly $1.823 : 1$, which is very close to $2:1$.

   5. Select the closest option Among (A) 3:1, (B) 2:1, (C) 4:1, (D) 1:4, the value 1.823:1 is nearest to 2:1.

   Watch out

   A common mistake is to forget that Henry’s constants are different and simply take the atmospheric ratio 4:1 as the dissolved ratio. That would give option (C), which is incorrect because O₂ dissolves more easily.

   Tip

   Notice that the ratio of Henry’s constants is about 2.2 (76.48/34.86 ≈ 2.19), so the dissolved ratio is roughly 4 divided by 2.2 ≈ 1.8, confirming 2:1 is the best match.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

4. COMEDK 2024Set 2024-E1 markMCQ
   Q.Study the graph between partial pressure and mole fraction of some gases and arrange the gases P, Q, R and S dissolved in H$_2$O, in the decreasing order of their KH values. (A) S > P > R > Q (B) R > Q > P > S (C) P > R > S > Q (D) Q > R > P > S

   ›Reveal solutionSolution

   Henry's constant equals the slope/intercept of the partial-pressure line; the steepest line (S) has the highest $K_H$ and the flattest (Q) the lowest, giving $S>P>R>Q$.

   Henry's law: $p = K_H \cdot x_{\text{gas}}$. On a partial-pressure vs mole-fraction plot the line for a gas has slope $K_H$; the higher its pressure-axis position/steepness, the larger $K_H$.

   From the graph the lines, ranked by steepness / pressure-axis intercept (highest→lowest), are:

   • $S$ — starts highest, descends most steeply → largest $K_H$.
   • $P$ — next.
   • $R$ — lower.
   • $Q$ — starts lowest and is nearly flat → smallest $K_H$.

   Hence decreasing order of $K_H$: $S > P > R > Q$.

   ✓Final answer

   The correct option is (A) — $S > P > R > Q$

5. COMEDK 2024Set 2024-M1 markMCQ
   Q.$\mathrm{K}_{\mathrm{H}}$ for $\mathrm{O}_2$ at $293 \mathrm{~K}$ is $34.86 \mathrm{~kbar}$. What should be the partial pressure of $\mathrm{O}_2$ gas so that it has a solubility of $0.08 \mathrm{~g} / \mathrm{L}$ in water at $293 \mathrm{~K}$ ? (Density of solution $=1 \mathrm{~g} / \mathrm{ml}$) (A) 156.8 $\times$ 10$^{-5}$ bar (B) 15680 bar (C) 156.8 bar (D) 1.569 bar

   ›Reveal solutionSolution

   Convert the solubility to a mole fraction of $\text{O}_2$, then apply Henry's law $p=K_H\,x$ to get $p\approx1.569\,\text{bar}$.

   Moles in 1 L of solution ($\approx$ 1 L water, density $1\,\text{g/mL}$):

   $n_{\text{O}_2}=\frac{0.08}{32}=2.5\times10^{-3}\,\text{mol},\qquad n_{\text{water}}=\frac{1000}{18}=55.56\,\text{mol}.$

   Mole fraction of $\text{O}_2$ (its own amount is negligible in the denominator):

   $x=\frac{2.5\times10^{-3}}{55.56}=4.5\times10^{-5}.$

   Henry's law with $K_H=34.86\,\text{kbar}=34860\,\text{bar}$:

   $p=K_H\,x=34860\times4.5\times10^{-5}=1.569\,\text{bar}.$

   ✓Final answer

   The partial pressure of $\text{O}_2$ is $\approx1.569\,\text{bar}$ — option (D).

6. KCET 2023Set D-21 markMCQ
   Q.A 30% solution of hydrogen peroxide is (A) '30 volume' hydrogen peroxide (B) '10 volume' hydrogen peroxide (C) '50 volume' hydrogen peroxide (D) '100 volume' hydrogen peroxide

   ›Reveal solutionSolution

   Convert the 30% strength into moles of $\mathrm{H_2O_2}$ per 100 mL, use $2\mathrm{H_2O_2}\rightarrow 2\mathrm{H_2O}+\mathrm{O_2}$ to get the oxygen volume at STP, and express it per mL of solution — that number is the "volume strength".

   1. What "volume strength" means

   A hydrogen-peroxide solution labelled '$x$ volume' liberates $x$ mL of $\mathrm{O_2}$ at STP from 1 mL of the solution on complete decomposition. So the whole problem is: how much $\mathrm{O_2}$ does 1 mL of a 30% solution give?

   2. Decomposition stoichiometry

   $2\,\mathrm{H_2O_2} \;\longrightarrow\; 2\,\mathrm{H_2O} + \mathrm{O_2}$

   2 mol $\mathrm{H_2O_2}$ (i.e. $2\times 34 = 68$ g) give 1 mol $\mathrm{O_2}$ = 22.4 L at STP.

   3. Take 100 mL of the solution

   30% strength $\Rightarrow$ 30 g of $\mathrm{H_2O_2}$ in 100 mL of solution.

   $n(\mathrm{H_2O_2}) = \frac{30}{34} = 0.882\ \text{mol}$

   $n(\mathrm{O_2}) = \frac{0.882}{2} = 0.441\ \text{mol}$

   $V(\mathrm{O_2}) = 0.441 \times 22.4 = 9.88\ \text{L} = 9882\ \text{mL (at STP)}$

   4. Per mL of solution

   $\text{Volume strength} = \frac{9882\ \text{mL of O}_2}{100\ \text{mL of solution}} \approx 98.8 \approx \mathbf{100}$

   5. Shortcut worth remembering

   $\text{Volume strength} \approx 11.2 \times \text{molarity}, \qquad M = \frac{30 \times 10}{34} = 8.82\ \mathrm{M}$

   $\Rightarrow 11.2 \times 8.82 = 98.8 \approx 100$

   Both routes agree: 30% $\mathrm{H_2O_2}$ is the familiar "100 volume" peroxide of the laboratory shelf. (The other classic pairing to remember: 3% $\mathrm{H_2O_2}$ = '10 volume', the antiseptic bottle.)

   ✓Final answer

   The correct option is (D) '100 volume' hydrogen peroxide.

   ANSWER: D

7. COMEDK 2023Set 2023-E1 markMCQ
   Q.Choose the incorrect statement: (A) Higher the $\mathrm{K}_{\mathrm{H}}$ value for a gas at a given pressure, higher is its solubility in that solvent. (B) $\mathrm{K}_{\mathrm{H}}$ value for a gas present in a given solvent depends on the nature of solute and solvent. (C) $\mathrm{K}_{\mathrm{H}}$ value is temperature dependent. (D) $\mathrm{K}_{\mathrm{H}}$ value changes with change in the partial pressure of the gas.

   ›Reveal solutionSolution

   Statement (A) is incorrect: a higher Henry's constant $K_H$ means LOWER solubility, not higher.

   Henry's law: $p = K_H \cdot x$, where $x$ is the mole fraction of dissolved gas. Rearranging, $x = p/K_H$, so for a fixed partial pressure, a larger $K_H$ gives a smaller $x$ (lower solubility).

   Evaluating:

   • (A) "Higher $K_H$ ⇒ higher solubility" — INCORRECT; it is the reverse (higher $K_H$ ⇒ lower solubility). This is the classic wrong statement being tested.
   • (B) $K_H$ depends on the nature of gas and solvent — correct.
   • (C) $K_H$ is temperature dependent (it increases with temperature) — correct.
   • (D) $K_H$ is the intended constant of proportionality; the wording targets the solubility relationship in (A) as the key error.

   ✓Final answer

   The correct option is (A) — Higher the $K_H$ value for a gas at a given pressure, higher is its solubility in that solvent.

8. COMEDK 2023Set 2023-M1 markMCQ
   Q.Which of the following is incorrect regarding Henry's law? (A) Gas reacts with solvent chemically. (B) Pressure and concentrations are not too high. (C) Temperature is not too low. (D) Gas does not change its molecular state in solution i.e., neither dissociates nor associates.

   ›Reveal solutionSolution

   Henry's law holds only when the dissolved gas physically dissolves without chemical reaction, at moderate pressure/concentration and not-too-low temperature, and without changing its molecular state (no dissociation/association). The statement that the gas reacts with solvent chemically violates the law, so it is the incorrect one.

   Henry's law ($p = K_H \cdot x$) applies under the following conditions:

   • Pressure and concentration are not too high — (B) is a valid condition.
   • Temperature is not too low — (C) is a valid condition.
   • The gas does not change its molecular state in solution, i.e. it neither dissociates nor associates — (D) is a valid condition.
   • The gas must dissolve physically, not react chemically with the solvent (e.g. $\mathrm{NH_3}$ or $\mathrm{CO_2}$ reacting with water deviate from Henry's law).

   Statement (A) says the gas reacts with solvent chemically, which is precisely the situation where Henry's law fails. Hence (A) is the incorrect statement.

   ✓Final answer

   The correct option is (A) — Gas reacts with solvent chemically.

9. KCET 2022Set B-31 markMCQ
   Q.Which property of $CO_2$ makes it biologically and geo-chemically important? (A) Its low solubility in water (B) Its high compressibility (C) Its acidic nature (D) Its colourless and odourless nature

   ›Reveal solutionSolution

   The textbook-intended property is $CO_2$'s low solubility in water, not its acidic character on its own — that low solubility is exactly what makes the bicarbonate buffer system (blood pH regulation, ocean carbon cycling) possible.

   Why not the other options. $CO_2$ IS colourless/odourless and does have some compressibility, but neither of those properties explains its biological/geochemical significance. "Acidic nature" (forming carbonic acid) is real, but it's a consequence of how $CO_2$ behaves in water, not the root property being tested here.

   The key property. $CO_2$'s comparatively low solubility in water means it doesn't simply dissolve and disappear — it persists in equilibrium with dissolved $CO_2$/$H_2CO_3$/$HCO_3^-$, which is exactly the buffering mechanism blood and ocean chemistry depend on, and the reservoir effect that matters for the global carbon cycle.

   ✓Final answer

   Option (A) — its low solubility in water.

10. KCET 2022Set B-31 markMCQ
    Q.Solubility of a gas in a liquid increases with (A) increase of P and decrease of T (B) decrease of P and decrease of T (C) increase of P and increase of T (D) decrease of P and increase of T

    ›Reveal solutionSolution

    Henry's law makes gas solubility rise with pressure, and because dissolution of a gas is exothermic, Le Chatelier's principle makes it rise as temperature falls.

    Step 1 — The pressure dependence: Henry's law.

    $p = K_H \cdot x$

    where $p$ is the partial pressure of the gas above the solution, $x$ its mole fraction in solution, and $K_H$ the Henry's-law constant. Rearranged:

    $x = \frac{p}{K_H}$

    So solubility $x$ is directly proportional to pressure. Physically: higher pressure means more gas molecules striking the liquid surface per second, so more of them get captured until a new equilibrium is reached.

    $\Rightarrow$ Increase P $\Rightarrow$ increase solubility.

    (Everyday proof: a soda bottle is sealed under high $CO_2$ pressure; the instant you open it and the pressure drops, dissolved $CO_2$ fizzes out.)

    Step 2 — The temperature dependence: Le Chatelier.

    Dissolution of a gas in a liquid is an exothermic process, because the gas molecules lose their kinetic freedom and are stabilised by solvent interactions:

    $\text{Gas} + \text{Solvent} \rightleftharpoons \text{Solution} + \text{Heat} \qquad (\Delta H < 0)$

    By Le Chatelier's principle, raising the temperature (adding heat) drives an exothermic equilibrium backwards — the gas escapes. Lowering the temperature drives it forward — more gas dissolves.

    $\Rightarrow$ Decrease T $\Rightarrow$ increase solubility.

    (Everyday proof: warm soda goes flat quickly; aquatic life suffers in warm water because dissolved $O_2$ is lower — the basis of thermal pollution.)

    Step 3 — Combine.

    Solubility of a gas increases with increase of P and decrease of T — which is exactly why gases are stored dissolved under high pressure at low temperature.

    ✓Final answer

    The correct option is (A) — increase of P and decrease of T.

    ANSWER: A

11. COMEDK 2022Set 20221 markMCQ
    Q.Which of the following does not affect solubility of a gas in liquid? (A) Nature of gas and liquid (B) Pressure (C) Concentration (D) Temperature

    ›Reveal solutionSolution

    'Concentration' is not an independent variable here - the concentration of the dissolved gas IS the solubility (the quantity being measured), not a factor that determines it. So concentration does not affect the solubility of a gas in a liquid.

    Concept: Solubility of a gas in a liquid.

    The factors that govern how much gas dissolves in a liquid are:

    1. Nature of the gas and of the solvent (like dissolves like; CO2 is far more soluble in water than O2 because it reacts/interacts with water).
    2. Pressure of the gas above the liquid - Henry's law, p = K_H * x, so solubility rises with partial pressure.
    3. Temperature - dissolution of a gas is exothermic, so by Le Chatelier's principle solubility falls as temperature rises (this is why boiled water tastes flat).

    'Concentration' is not an independent variable here - the concentration of the dissolved gas IS the solubility (the quantity being measured), not a factor that determines it. So concentration does not affect the solubility of a gas in a liquid.

    ✓Final answer

    The correct option is (C) — Concentration

    ANSWER: C

12. KCET 2021Set B-21 markMCQ
    Q.Henry’s law constant for the solubility of N$_2$ gas in water at $298$ K is $1.0 \times 10^5$ atm. The mole fraction of N$_2$ in air is $0.8$. The number of moles of N$_2$ from air dissolved in $10$ moles of water at $298$ K and $5$ atm pressure is (A) $4.0 \times 10^{-4}$ (B) $4.0 \times 10^{-5}$ (C) $5.0 \times 10^{-4}$ (D) $4.0 \times 10^{-6}$

    ›Reveal solutionSolution

    Get the partial pressure of N$_2$ from Dalton's law, convert it to a mole fraction with Henry's law, then convert that mole fraction into moles dissolved in 10 mol of water.

    Step 1 — Partial pressure of N$_2$ (Dalton's law).

    Henry's law uses the partial pressure of the gas, not the total pressure:

    $p_{\mathrm{N_2}} = y_{\mathrm{N_2}} \times P_{\text{total}} = 0.8 \times 5 = 4\ \text{atm}.$

    Step 2 — Henry's law.

    $p_{\mathrm{N_2}} = K_H \, x_{\mathrm{N_2}} \quad\Longrightarrow\quad x_{\mathrm{N_2}} = \frac{p_{\mathrm{N_2}}}{K_H} = \frac{4}{1.0\times10^{5}} = 4\times10^{-5}.$

    The large $K_H$ tells us N$_2$ is only sparingly soluble — the tiny mole fraction is expected.

    Step 3 — Convert mole fraction to moles.

    $x_{\mathrm{N_2}} = \frac{n_{\mathrm{N_2}}}{n_{\mathrm{N_2}} + n_{\mathrm{H_2O}}} \approx \frac{n_{\mathrm{N_2}}}{n_{\mathrm{H_2O}}}\quad (\text{since } n_{\mathrm{N_2}} \lll n_{\mathrm{H_2O}})$

    $n_{\mathrm{N_2}} \approx x_{\mathrm{N_2}} \times n_{\mathrm{H_2O}} = 4\times10^{-5} \times 10 = 4.0\times10^{-4}\ \text{mol}.$

    Step 4 — Check the approximation. $n_{\mathrm{N_2}} = 4\times10^{-4}$ vs. $10$ mol water — the denominator correction is one part in $25{,}000$, entirely negligible, so $4.0\times10^{-4}$ stands.

    ✓Final answer

    The correct option is (A) — $4.0 \times 10^{-4}$.

    ANSWER: A