Q.Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given below: 2NO

Showing the 12 most recent of 49 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.A gaseous mixture contains H$_2$ and O$_2$. The pressure of the mixture is 1 bar. The weight percentage of O$_2$ is 80. The ratio of partial pressures of H$_2$ and O$_2$ is (A) 5 (B) 4 (C) 0.2 (D) 0.25

   ›Reveal solutionSolution

   Take $100\text{ g}$ of mixture: $80\text{ g }O_2$ and $20\text{ g }H_2$; the partial-pressure ratio equals the mole ratio $= 4$.

   Weight percentage of $O_2$ is $80\%$, so in $100\text{ g}$ of mixture there are $80\text{ g }O_2$ and $20\text{ g }H_2$.

   $n_{O_2} = \frac{80}{32} = 2.5\text{ mol},\qquad n_{H_2} = \frac{20}{2} = 10\text{ mol}$ …

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2. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.The following equilibrium is established at 1100 K in a closed V L flask C(s) + CO$_2$(g) $\rightleftharpoons$ 2CO(g) The pressure of equilibrium mixture is 1 atm. Among the gaseous compounds, CO has 84% by mass. $K_c$ of this reaction is (approximately) (R = 0.082 L atm mol$^{-1}$K$^{-1}$) (Assume that CO and CO$_2$ are ideal gases) (A) $10 \times 10^{-2}$ (B) $3 \times 10^{-2}$ (C) $1 \times 10^{-2}$ (D) $8 \times 10^{-2}$

   ›Reveal solutionSolution

   The key is to use the mass percentage of CO to find the mole fractions and partial pressures, then convert $K_p$ to $K_c$ via $K_c = K_p (RT)^{-\Delta n}$. The approximate value is $3 \times 10^{-2}$.

   The reaction involves a solid (carbon) and two gases. Since the solid’s activity is constant, the equilibrium constant in terms of partial pressures is $K_p = \frac{P_{\text{CO}}^2}{P_{\text{CO}_2}}$. We are given the total pressure (1 atm) and the mass composition of the gas mixture — CO is 84% by mass. That means in every 100 g of gas mixture, 84 g is CO and 16 g is CO$_2$. From mass, we can get moles, then mole fractions, then partial pressures, and finally $K_p$. The question asks for $K_c$, so we’ll use the relation $K_p = K_c (RT)^{\Delta n}$, where $\Delta n = 2 - 1 = 1$.

   1. Find the mole ratio from mass data.

      Molar mass of CO = 28 g/mol, of CO$_2$ = 44 g/mol.

      In 100 g of mixture:

      Moles of CO = $\frac{84}{28} = 3$ mol

      Moles of CO$_2$ = $\frac{16}{44} = \frac{4}{11} \approx 0.3636$ mol

      Total moles of gas = $3 + \frac{4}{11} = \frac{33}{11} + \frac{4}{11} = \frac{37}{11} \approx 3.3636$ mol

   2. Mole fractions and partial pressures.

      Mole fraction of CO: $y_{\text{CO}} = \frac{3}{37/11} = \frac{33}{37}$

      Mole fraction of CO$_2$: $y_{\text{CO}_2} = \frac{4/11}{37/11} = \frac{4}{37}$

      Total pressure $P = 1$ atm, so:

      $P_{\text{CO}} = y_{\text{CO}} \times 1 = \frac{33}{37}$ atm

      $P_{\text{CO}_2} = y_{\text{CO}_2} \times 1 = \frac{4}{37}$ atm

   3. Calculate $K_p$.

      $K_p = \frac{P_{\text{CO}}^2}{P_{\text{CO}_2}} = \frac{(33/37)^2}{4/37} = \frac{33^2}{37^2} \times \frac{37}{4} = \frac{1089}{37 \times 4} = \frac{1089}{148}$

      Simplify: $1089 \div 148 \approx 7.358$

      So $K_p \approx 7.36$ atm.

   4. Convert $K_p$ to $K_c$.

      For the reaction $\text{C(s)} + \text{CO}_2(g) \rightleftharpoons 2\text{CO}(g)$, $\Delta n = 2 - 1 = 1$. …

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3. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.At T(K) consider the following equilibrium reaction HgO (s) $\rightleftharpoons$ Hg

   (g) + $\frac{1}{2}$O$_2$(g) The correct relation between $K_p$ and $P_\text{Total}$ ($P_T$) is (A) $K_p = \frac{2}{3^{1/2}} \cdot P_T^{1/2}$ (B) $K_p = \frac{2}{3^{3/2}} \cdot P_T^{1/2}$ (C) $K_p = \frac{2}{3^{3/2}} \cdot P_T^{3/2}$ (D) $K_p = \frac{1}{3^{3/2}} \cdot P_T$

   ›Reveal solutionSolution

   For the decomposition of solid HgO, the partial pressures of the gaseous products Hg and O$_2$ are related by their stoichiometry. Expressing these partial pressures in terms of the total pressure $P_T$ and substituting into the $K_p$ expression yields $K_p = \frac{2}{3^{3/2}} P_T^{3/2}$.

   The equilibrium constant $K_p$ for a reaction involving gases is defined in terms of the partial pressures of the gaseous reactants and products. For the given reaction, HgO (s) $\rightleftharpoons$ Hg (g) + $\frac{1}{2}$O$_2$(g), only the gaseous species, Hg(g) and O$_2$(g), contribute to the $K_p$ expression. The solid reactant, HgO(s), does not appear in the $K_p$ expression because its "concentration" (or activity) is considered constant and is absorbed into the value of $K_p$.

   The key idea here is to relate the partial pressures of the gaseous products to the total pressure of the system using their stoichiometric coefficients. Since both Hg(g) and O$_2$(g) are produced from the decomposition of HgO(s), their relative amounts (and thus their partial pressures) will be directly proportional to their stoichiometric coefficients in the balanced chemical equation.

   1. Identify Gaseous Species and Stoichiometry:

      The equilibrium reaction is:

      HgO (s) $\rightleftharpoons$ Hg (g) + $\frac{1}{2}$O$_2$(g)

      The gaseous products are Hg(g) and O$_2$(g).

      From the stoichiometry, for every 1 mole of Hg(g) produced, $\frac{1}{2}$ mole of O$_2$(g) is produced. This means the number of moles of O$_2$ is half the number of moles of Hg.

   2. Relate Partial Pressures to Stoichiometry:

      According to Dalton's Law of Partial Pressures, the partial pressure of a gas in a mixture is proportional to its mole fraction. Since the volume and temperature are common for all gases, the ratio of partial pressures is equal to the ratio of their moles.

      Let $P_{\text{Hg}}$ be the partial pressure of Hg(g) and $P_{\text{O}_2}$ be the partial pressure of O$_2$(g).

      Based on the stoichiometry:

      $P_{\text{O}_2} = \frac{1}{2} P_{\text{Hg}}$

   3. Express Partial Pressures in Terms of Total Pressure ($P_T$):

      The total pressure $P_T$ is the sum of the partial pressures of all gaseous components:

      $P_T = P_{\text{Hg}} + P_{\text{O}_2}$

      Substitute the relationship from step 2 into this equation:

      $P_T = P_{\text{Hg}} + \frac{1}{2} P_{\text{Hg}}$

      $P_T = \left(1 + \frac{1}{2}\right) P_{\text{Hg}}$

      $P_T = \frac{3}{2} P_{\text{Hg}}$

      Now, we can express $P_{\text{Hg}}$ in terms of $P_T$:

      $P_{\text{Hg}} = \frac{2}{3} P_T$

      And subsequently, $P_{\text{O}_2}$ in terms of $P_T$: …

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4. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If the kinetic energy of the molecules in 8.0 g methane at 47°C is $x$ kJ, then kinetic energy (in kJ) of the molecules in same amount of dihydrogen at 127°C is (A) $x$ (B) $5x$ (C) $10x$ (D) $20x$

   ›Reveal solutionSolution

   The kinetic energy of an ideal gas depends only on the number of molecules and the absolute temperature. For equal masses, the ratio of kinetic energies is the ratio of mole counts times the ratio of absolute temperatures. The answer is $10x$.

   The key idea here is that the average kinetic energy of a gas molecule depends only on temperature — not on the type of molecule. For an ideal gas, the total kinetic energy of a sample is simply the number of molecules times the average kinetic energy per molecule. That means if you have the same number of molecules at the same temperature, the total kinetic energy is identical, regardless of whether the gas is methane or hydrogen.

   But here the masses are equal, not the mole counts. So the first step is to figure out how many molecules (or moles) are present in each case. Then apply the temperature difference.

   Let’s work through it step by step.

   1. Find the number of moles in 8.0 g of each gas.

      Molar mass of methane ($CH_4$) = $12 + 4 \times 1 = 16$ g/mol.

      Moles of methane = $\frac{8.0}{16} = 0.5$ mol.

      Molar mass of dihydrogen ($H_2$) = $2 \times 1 = 2$ g/mol.

      Moles of dihydrogen = $\frac{8.0}{2} = 4$ mol.

      So the same mass gives 8 times more moles of hydrogen than methane.

   2. Recall the formula for total kinetic energy of an ideal gas.

      For $n$ moles at absolute temperature $T$, the total translational kinetic energy is

   $KE = \frac{3}{2} nRT$

   where $R$ is the universal gas constant. This comes from the kinetic theory: each molecule has $\frac{3}{2}k_BT$ on average, and there are $nN_A$ molecules.

   3. Write the kinetic energy for methane. Methane: $n_1 = 0.5$ mol, $T_1 = 47^\circ C = 47 + 273 = 320$ K.

   $KE_{\text{methane}} = \frac{3}{2} \times 0.5 \times R \times 320 = \frac{3}{2} \times 160 R = 240R$

   And we are told this equals $x$ kJ. So $x = 240R$ (in kJ, if $R$ is in kJ/mol·K).

   4. Write the kinetic energy for dihydrogen. …

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5. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.At 300 K, a 10 L vessel contains 0.4 g of He, 1.6 g of O$_2$ and 1.4 g of N$_2$. The partial pressure of He gas (in atm) is (Assume ideal behaviour for all gases) (R = 0.0821 L atm K$^{-1}$ mol$^{-1}$) (A) 0.123 (B) 0.246 (C) 0.323 (D) 0.133

   ›Reveal solutionSolution

   The partial pressure of He is found by calculating its moles, then using the ideal gas law with the given volume and temperature. The result is 0.246 atm, which corresponds to option (B).

   Concept & Intuition

   In a mixture of ideal gases, each gas behaves as if it alone occupies the container. The partial pressure of a gas is the pressure it would exert if it were the only gas present, at the same temperature and volume. So we don’t need the other gases at all — just the moles of He, the volume, and the temperature. The ideal gas law $PV = nRT$ directly gives the partial pressure.

   Step-by-step solution

   1. Find the number of moles of He Molar mass of He = 4 g/mol. Mass of He = 0.4 g.

   $n_{\text{He}} = \frac{0.4\ \text{g}}{4\ \text{g/mol}} = 0.1\ \text{mol}$

   2. Apply the ideal gas law for He alone The partial pressure $P_{\text{He}}$ satisfies

   $P_{\text{He}} V = n_{\text{He}} R T$

   Given: $V = 10\ \text{L}$, $T = 300\ \text{K}$, $R = 0.0821\ \text{L atm K}^{-1}\text{mol}^{-1}$.

   $P_{\text{He}} = \frac{n_{\text{He}} R T}{V} = \frac{0.1 \times 0.0821 \times 300}{10}$

   3. Calculate First, $0.1 \times 0.0821 = 0.00821$. …

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6. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.At T(K), the kinetic energy of one mole of an ideal gas is 3735 J. If its pressure is 1 atm, the volume of gas is ($R = 8.3~\mathrm{J~mol^{-1}~K^{-1}}$) (A) 12.3 L (B) 24.6 L (C) 49.2 L (D) 36.8 L

   ›Reveal solutionSolution

   The kinetic energy of one mole of an ideal gas is directly related to its temperature. By first calculating the temperature from the given kinetic energy, we can then use the ideal gas law to find the volume. The volume of the gas is $\boxed{\text{24.6 L}}$.

   The kinetic energy of an ideal gas is a direct measure of its absolute temperature. For an ideal gas, the internal energy is purely kinetic, and this kinetic energy is distributed among the translational degrees of freedom of the gas molecules. The average translational kinetic energy per molecule is $\frac{3}{2}kT$, where $k$ is Boltzmann's constant. For one mole of gas, we multiply this by Avogadro's number ($N_A$), and since $k N_A = R$ (the ideal gas constant), the total translational kinetic energy for one mole of an ideal gas is $\frac{3}{2}RT$.

   Once we determine the temperature of the gas using this relationship, we can then use the ideal gas law, $PV = nRT$, to find the volume. This law connects the macroscopic properties of pressure ($P$), volume ($V$), temperature ($T$), and the number of moles ($n$) of an ideal gas.

   Here's how to solve the problem step-by-step:

   1. Calculate the temperature (T) using the kinetic energy:

      The kinetic energy of one mole of an ideal gas is given by the formula:

      $E_k = \frac{3}{2}RT$

      We are given $E_k = 3735~\mathrm{J}$ and $R = 8.3~\mathrm{J~mol^{-1}~K^{-1}}$. We can rearrange the formula to solve for $T$:

      $T = \frac{2 E_k}{3R}$

      Substitute the given values:

      $T = \frac{2 \times 3735~\mathrm{J}}{3 \times 8.3~\mathrm{J~mol^{-1}~K^{-1}}}$

      $T = \frac{7470}{24.9}~\mathrm{K}$

      $T = 300~\mathrm{K}$

   2. Convert pressure to SI units:

      The ideal gas law ($PV=nRT$) requires pressure in Pascals (Pa) if $R$ is in $\mathrm{J~mol^{-1}~K^{-1}}$ and volume is in $\mathrm{m^3}$.

      We are given $P = 1~\mathrm{atm}$. We know that $1~\mathrm{atm} = 1.01325 \times 10^5~\mathrm{Pa}$.

      So, $P = 1.01325 \times 10^5~\mathrm{Pa}$. …

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7. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.At 293 K, HCl

   (g) is dissolved in cyclohexane. Identify the graph obtained for this solution (x-axis = mole fraction of HCl dissolved in cyclohexane; y-axis = partial pressure of HCl (g)) (A) [FIGURE] A straight line of positive slope starting from a positive intercept on the y-axis (does not pass through the origin) (B) [FIGURE] A straight line of negative slope, starting high on the y-axis and falling towards the x-axis (C) [FIGURE] A straight line of positive slope passing through the origin (D) [FIGURE] A horizontal straight line (y constant, independent of x)

   ›Reveal solutionSolution

   HCl dissolved in cyclohexane behaves as a simple Henry's-law solute, so $p_{\mathrm{HCl}} = K_H\, x_{\mathrm{HCl}}$. This is a straight line of positive slope passing exactly through the origin — option (C).

   The concept first

   When a gas dissolves physically (no reaction, no ionisation) in a liquid, Henry's law applies:

   $p = K_H \, x$

   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 at that temperature. Notice the law is linear and homogeneous: no additive constant.

   Step-by-step

   1. Identify the system. HCl(g) in cyclohexane at $293\ \mathrm{K}$. Cyclohexane is non-polar; HCl does not ionise in it and forms no compound with it. So HCl is simply a dissolved gas — a Henry's-law solute.
   2. Write the relation. $p_{\mathrm{HCl}} = K_H\, x_{\mathrm{HCl}}$, with $K_H$ constant at the fixed temperature $293\ \mathrm{K}$.
   3. Read it as a graph. With $y = p_{\mathrm{HCl}}$ and $x = x_{\mathrm{HCl}}$, this is $y = K_H x$, i.e. $y = mx$ with $m = K_H > 0$: a straight line, positive slope, intercept zero. …

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8. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.For the reaction $\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}$, the correct relation between degree of dissociation $(\alpha)$ of $\mathrm{N_2O_4(g)}$ and equilibrium constant, $K_p$ is (P = total pressure of mixture) (A) $\alpha = \dfrac{\dfrac{K_p}{P}}{4 + \dfrac{K_p}{P}}$ (B) $\alpha = \dfrac{K_p}{4 + K_p}$ (C) $\alpha = \left( \dfrac{\dfrac{K_p}{P}}{4 + \dfrac{K_p}{P}} \right)^{\frac{1}{2}}$ (D) $\alpha = \left( \dfrac{K_p}{4 + K_p} \right)^{\frac{1}{2}}$

   ›Reveal solutionSolution

   The key is to express the equilibrium constant $K_p$ in terms of the degree of dissociation $\alpha$ and total pressure $P$, then solve for $\alpha$. The correct relation is $\alpha = \sqrt{\frac{K_p/P}{4 + K_p/P}}$, which corresponds to option (C).

   ---

   The reaction is $\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}$.

   We start with 1 mole of $\mathrm{N_2O_4}$ and let $\alpha$ be the fraction that dissociates.

   The total number of moles at equilibrium changes, so partial pressures depend on both $\alpha$ and total pressure $P$.

   The equilibrium constant $K_p$ is defined in terms of partial pressures, so we need to express each partial pressure as (mole fraction) × (total pressure).

   ---

   1. Set up the initial and equilibrium moles

      • Initial: $\mathrm{N_2O_4} = 1$ mole, $\mathrm{NO_2} = 0$
      • At equilibrium: $\mathrm{N_2O_4}$ remaining = $1 - \alpha$ $\mathrm{NO_2}$ formed = $2\alpha$
      • Total moles at equilibrium: $n_{\text{total}} = (1 - \alpha) + 2\alpha = 1 + \alpha$

   2. Write mole fractions and partial pressures

      • Mole fraction of $\mathrm{N_2O_4}$: $\dfrac{1 - \alpha}{1 + \alpha}$ Partial pressure $P_{\mathrm{N_2O_4}} = \dfrac{1 - \alpha}{1 + \alpha} \cdot P$
      • Mole fraction of $\mathrm{NO_2}$: $\dfrac{2\alpha}{1 + \alpha}$ Partial pressure $P_{\mathrm{NO_2}} = \dfrac{2\alpha}{1 + \alpha} \cdot P$

   3. Write the expression for $K_p$

      $K_p = \dfrac{(P_{\mathrm{NO_2}})^2}{P_{\mathrm{N_2O_4}}}$

      Substitute:

   $K_p = \frac{\left( \dfrac{2\alpha}{1 + \alpha} \cdot P \right)^2}{\dfrac{1 - \alpha}{1 + \alpha} \cdot P} = \frac{4\alpha^2 P^2}{(1 + \alpha)^2} \cdot \frac{1 + \alpha}{(1 - \alpha)P} = \frac{4\alpha^2 P}{(1 + \alpha)(1 - \alpha)}$

   4. Simplify the denominator $(1 + \alpha)(1 - \alpha) = 1 - \alpha^2$ So:

   $K_p = \frac{4\alpha^2 P}{1 - \alpha^2}$

   5. Solve for $\alpha$ Multiply both sides: $K_p (1 - \alpha^2) = 4\alpha^2 P$ $\Rightarrow K_p - K_p \alpha^2 = 4P \alpha^2$ …

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9. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.At 1000 K, the equilibrium constant for the reaction $\mathrm{CO_2(g) + H_2(g) \rightleftharpoons CO(g) + H_2O(g)}$ is 0.53. In a one litre vessel, at equilibrium the mixture contains 0.25 mole of CO, 0.5 mole of CO$_2$, 0.6 mole of H$_2$ and $x$ moles of H$_2$O. The value of $x$ is (A) 0.563 (B) 0.363 (C) 0.636 (D) 0.736

   ›Reveal solutionSolution

   Apply $K_c=\dfrac{[\text{CO}][\text{H}_2\text{O}]}{[\text{CO}_2][\text{H}_2]}$ and solve for $x$; this gives $x = 0.636$ — option (C).

   Concept

   For $\text{CO}_2(g)+\text{H}_2(g)\rightleftharpoons \text{CO}(g)+\text{H}_2\text{O}(g)$ the equilibrium constant is

   $K_c = \frac{[\text{CO}]\,[\text{H}_2\text{O}]}{[\text{CO}_2]\,[\text{H}_2]}.$

   Since the vessel is $1$ L, each molar concentration equals the number of moles.

   Solution …

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10. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
    Q.The osmotic pressure (in atm) of an aqueous solution containing 0.01 mol of NaCl (degree of dissociation 0.94) and 0.03 mol of glucose in 500 mL at 27 $^\circ$C is (R = 0.082 L atm K$^{-1}$mol$^{-1}$) (A) 2.43 (B) 4.23 (C) 3.24 (D) 3.42

    ›Reveal solutionSolution

    Sum the effective particles: NaCl ($i = 1.94$) plus glucose ($i = 1$), then apply $\Pi = \dfrac{n_{\text{eff}}RT}{V}$ to get $2.43$ atm — option (A).

    Concept

    Osmotic pressure counts total dissolved particles: $\Pi V = n_{\text{eff}}RT$, where for NaCl the van't Hoff factor is $i = 1 + \alpha(n-1)$.

    Solution

    NaCl ($n = 2$, $\alpha = 0.94$): $i = 1 + 0.94(2-1) = 1.94$.

    Effective particles: …

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11. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
    Q.At 298K, a flask ‘A’ of unknown volume (V) contains oxygen at 5 atm. Another flask ‘B’ of volume 2L contains helium at 3 atm. Two flasks are connected together by a small tube of zero volume. After the two gases are completely mixed, if the resulting mixture is found to have the mole fraction of oxygen as 0.2, the volume of flask ‘A’ (in L) is (Assume oxygen and helium as ideal gases) (A) 0.1 (B) 0.3 (C) 0.2 (D) 0.4

    ›Reveal solutionSolution

    Setting the mole fraction of oxygen to 0.2 gives $V = 0.3~\text{L}$ — option (B).

    Concept

    At constant temperature the moles of an ideal gas are proportional to $PV$, so the factor $1/RT$ cancels and mole fractions can be built directly from $PV$ products.

    Solution

    Moles (as $PV$). Oxygen in flask A: $n_{\text{O}_2} \propto 5V$. Helium in flask B: $n_{\text{He}} \propto 3\times 2 = 6$.

    Mole fraction of oxygen after mixing:

    $x_{\text{O}_2} = \frac{5V}{5V + 6} = 0.2$

    Solve for $V$: …

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12. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
    Q.At 298K, a flask ‘A’ of unknown volume (V) contains oxygen at 5 atm. Another flask ‘B’ of volume 2L contains helium at 3 atm. Two flasks are connected together by a small tube of zero volume. After the two gases are completely mixed, if the resulting mixture is found to have the mole fraction of oxygen as 0.2, the volume of flask ‘A’ (in L) is (Assume oxygen and helium as ideal gases) (A) 0.2 (B) 0.3 (C) 0.4 (D) 0.1

    ›Reveal solutionSolution

    The key idea is that the mole fraction of oxygen after mixing equals the ratio of its initial moles to the total moles. Using the ideal gas law, we set up an equation and solve for the unknown volume V, finding V = 0.4 L.

    Concept and Intuition

    When two gases mix at constant temperature, the total number of moles is conserved. The mole fraction of a component after mixing is simply the number of moles of that component divided by the total moles. Since both gases are ideal and at the same temperature (298 K), we can use $PV = nRT$ to relate pressure, volume, and moles. The key is that the initial pressures and volumes give us the moles of each gas, and the final mixture’s mole fraction of oxygen provides an equation to solve for the unknown volume V.

    Step-by-step solution

    1. Find moles of oxygen in flask A

       For oxygen: pressure $P_{O_2} = 5\ \text{atm}$, volume $V_A = V\ \text{L}$, temperature $T = 298\ \text{K}$.

       Using the ideal gas law: $n_{O_2} = \frac{P_{O_2} V_A}{RT} = \frac{5V}{RT}$.

    2. Find moles of helium in flask B

       For helium: pressure $P_{He} = 3\ \text{atm}$, volume $V_B = 2\ \text{L}$.

       So $n_{He} = \frac{P_{He} V_B}{RT} = \frac{3 \times 2}{RT} = \frac{6}{RT}$.

    3. Total moles after mixing

       The total moles in the mixture:

    $n_{\text{total}} = n_{O_2} + n_{He} = \frac{5V}{RT} + \frac{6}{RT} = \frac{5V + 6}{RT}.$

    4. Use the given mole fraction of oxygen The mole fraction of oxygen after mixing is 0.2:

    $\frac{n_{O_2}}{n_{\text{total}}} = 0.2.$

    Substitute the expressions:

    $\frac{\frac{5V}{RT}}{\frac{5V + 6}{RT}} = 0.2.$

    The $RT$ cancels, giving: …

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