Q.Calculate the mole fraction of gas A in the solution made on mixing 0.5 moles of gas A and 4.5 moles of gas B.

Showing the 12 most recent of 18 on this concept.

1. CBSE 2026Set V11 markMCQ
   Q.A binary solution has two components 'A' and 'B'. The mole fraction of components 'A' is 0.5, then the number of moles of components 'A' and 'B' in the solution is

   (a) $n_A > n_B$

   (b) $n_A < n_B$

   (c) $n_A = n_B$

   (d) zero

   ›Reveal solutionSolution

   Equal mole fractions (0.5 each) mean equal moles of A and B.

   For a binary solution of components A and B, the mole fractions must add up to 1:

   $x_A + x_B = 1$

   Given $x_A = 0.5$, therefore $x_B = 1 - 0.5 = 0.5$.

   Since $x_A = \dfrac{n_A}{n_A + n_B}$ and $x_B = \dfrac{n_B}{n_A + n_B}$, and $x_A = x_B$, it follows that: …

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2. CBSE 2025Set A1 mark
   Q.Fill in the blank: The number of moles of solute dissolved in one litre (or one cubic decimeter) of solution is defined as ______.

   ›Reveal solutionSolution

   This blank describes the definition of molarity.

   Molarity (M) is one of the standard ways to express the concentration of a solution. It is defined as:

   $M = \dfrac{\text{moles of solute}}{\text{volume of solution in litres}}$

   …

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3. CBSE 2025Set ANNUAL1 markMCQ
   Q.What is the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride?

   (a) 0.540

   (b) 0.459

   (c) 0.500

   (d) 0.300

   ›Reveal solutionSolution

   Taking 100 g of solution (30 g benzene + 70 g CCl₄) and converting to moles gives a benzene mole fraction of 0.459.

   Assume 100 g of solution: mass of benzene ($C_6H_6$, M = 78 g mol⁻¹) = 30 g; mass of $CCl_4$ (M = 12 + 4×35.5 = 154 g mol⁻¹) = 70 g.

   $n_{benzene} = \frac{30}{78} = 0.3846\ mol$

   $n_{CCl_4} = \frac{70}{154} = 0.4545\ mol$

   …

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4. CBSE 2025Set ANNUAL1 markMCQ
   Q.1 mole of NaOH is present in 250 mL of solution, the molarity of solution in mol L^-1 will be -

   (a) 4

   (b) 40

   (c) 25

   (d) 0.4

   ›Reveal solutionSolution

   Molarity = moles of solute ÷ volume of solution (in litres); here 1 mol NaOH in 250 mL solution gives a molarity of 4 mol L^-1.

   Molarity (M) is defined as the number of moles of solute dissolved per litre of solution:

   M = (moles of solute) / (volume of solution in L)

   …

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5. CBSE 2024Set FZ1 markMCQ
   Q.5 g NaOH is dissolved in 450 mL solution. Molarity of the solution is:

   (a) $0.125\ \text{mol L}^{-1}$

   (b) $0.139\ \text{mol L}^{-1}$

   (c) $0.250\ \text{mol L}^{-1}$

   (d) $0.278\ \text{mol L}^{-1}$

   ›Reveal solutionSolution

   Moles of NaOH $=5/40=0.125$; molarity $=0.125/0.450=0.278\ \text{mol L}^{-1}$ → option (d).

   Concept. Molarity is the number of moles of solute dissolved per litre of solution: $M=\dfrac{n}{V(\text{L})}$.

   Step 1 — moles of NaOH. Molar mass of NaOH $=23+16+1=40\ \text{g mol}^{-1}$.

   $n=\dfrac{5}{40}=0.125\ \text{mol}$

   …

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6. CBSE 2024Set B1 markMCQ
   Q.5 gram NaOH is dissolved in 250 ml of solution. The molarity of solution will be

   (a) 0.5 M

   (b) 0.25 M

   (c) 2.5 M

   (d) 5.0 M

   ›Reveal solutionSolution

   Molarity = moles of solute divided by volume of solution in litres; here 0.125 mol NaOH in 0.250 L gives 0.5 M.

   Molar mass of NaOH = 23 (Na) + 16 (O) + 1 (H) = 40 g/mol.

   Moles of NaOH:

   $n = \frac{\text{given mass}}{\text{molar mass}} = \frac{5\text{ g}}{40\text{ g/mol}} = 0.125\text{ mol}$

   …

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7. CBSE 2024Set ANNUAL1 mark
   Q.The unit of molarity is ______.

   ›Reveal solutionSolution

   Molarity is defined as moles of solute dissolved per litre of solution, so its unit is mol per litre.

   Molarity (M) = number of moles of solute / volume of solution in litres. …

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8. CBSE 2024Set ANN1 mark
   Q.The number of moles of solute in one litre of the solution is ________.

   (a) Molarity

   (b) Molality

   (c) Normality

   (d) Mole fraction

   ›Reveal solutionSolution

   Molarity is the number of moles of solute dissolved per litre of solution.

   Concentration of a solution can be expressed in several ways — molarity, molality, normality and mole fraction — and each has its own defining ratio.

   Molarity (M) is defined as:

   M = (Number of moles of solute) / (Volume of solution in litres)

   This is different from:

   • Molality — moles of solute per kilogram of solvent.
   • Normality — gram equivalents of solute per litre of solution. …

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9. CBSE 2024Set ANNUAL1 markMCQ
   Q.The number of moles of Solute per litre of the Solution is known as:

   (a) Molality

   (b) Molarity

   (c) Normality

   (d) Formality

   ›Reveal solutionSolution

   Molarity (M) is the number of moles of solute per litre of solution; it is the most commonly used concentration unit in the lab.

   Concentration terms differ in what they divide moles (or mass) by:

   • Molality (m) = moles of solute per kilogram of solvent (mass-based, temperature-independent).
   • Molarity (M) = moles of solute per litre of solution — this is exactly what the question describes.
   • Normality (N) = number of gram-equivalents of solute per litre of solution. …

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10. CBSE 2023Set A1 mark
    Q.Fill in the blank: Molarity of water is ______.

    ›Reveal solutionSolution

    Pure water's molarity is found by dividing the mass of 1 litre of water by its molar mass.

    1 litre of water has a mass of approximately 1000 g (density ≈ 1 g/mL).

    Molar mass of water (H2O) = 18 g mol⁻¹

    …

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11. CBSE 2023Set ANNUAL1 mark
    Q.Define molarity of a solution.

    ›Reveal solutionSolution

    Molarity is the number of moles of solute dissolved per litre of solution.

    Molarity (M) is defined as the number of moles of solute present in one litre (dm3) of solution.

    Molarity (M) = (Number of moles of solute) / (Volume of solution in litres)

    …

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12. CBSE 2022Set GH1 markMCQ
    Q.7 g NaOH is dissolved in 400 mL solution. The molarity of solution is:

    (a) $0.437\ \text{mol L}^{-1}$

    (b) $0.496\ \text{mol L}^{-1}$

    (c) $0.546\ \text{mol L}^{-1}$

    (d) $0.637\ \text{mol L}^{-1}$

    ›Reveal solutionSolution

    Molarity $=\dfrac{\text{moles of solute}}{\text{volume of solution in litres}}=\dfrac{0.175}{0.400}=0.437\ \text{mol L}^{-1}$.

    Concept — molarity is the number of moles of solute dissolved per litre of solution: $M=\dfrac{n}{V(\text{L})}$. First convert the mass of NaOH to moles, then divide by the volume in litres.

    Step 1 — moles of NaOH (molar mass $=23+16+1=40\ \text{g mol}^{-1}$):

    $n=\frac{7\ \text{g}}{40\ \text{g mol}^{-1}}=0.175\ \text{mol}$

    …

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