Q.Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.

Showing the 12 most recent of 31 on this concept.

1. CBSE 2026Set 56/3/11 markMCQ
   Q.The mole fraction of a solute in $2.0$ molal aqueous solution is : (A) $1.87$ (B) $0.347$ (C) $0.0347$ (D) $0.00347$

   ›Reveal solutionSolution

   A 2.0 molal solution means 2 moles of solute in 1 kg of water. Convert the solvent mass to moles, then apply the mole fraction formula: $\chi_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{solvent}}}$. The answer is $\boxed{0.0347}$.

   Molality is defined as moles of solute per kilogram of solvent, not per kilogram of solution. This distinction matters because we need to count the moles of both solute and solvent separately to find the mole fraction.

   When we say a solution is 2.0 molal, we're saying there are 2.0 moles of solute dissolved in exactly 1000 g (1 kg) of water. The mole fraction then asks: what fraction of the total number of particles (molecules) in the solution comes from the solute?

   Let me work through the calculation systematically.

   1. Identify what we know from "2.0 molal aqueous solution"

   The molality $m = 2.0$ tells us:

   • Moles of solute: $n_{\text{solute}} = 2.0 \text{ mol}$
   • Mass of water (solvent): $1000 \text{ g}$

   2. Convert the mass of water to moles

   Water has a molar mass of $18 \text{ g/mol}$, so:

   $n_{\text{water}} = \frac{1000 \text{ g}}{18 \text{ g/mol}} = 55.56 \text{ mol}$

   3. Calculate the total moles in the solution

   $n_{\text{total}} = n_{\text{solute}} + n_{\text{water}} = 2.0 + 55.56 = 57.56 \text{ mol}$

   4. Apply the mole fraction formula

   The mole fraction of solute is:

   $\chi_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{total}}} = \frac{2.0}{57.56} = 0.03474$

   Rounding to three significant figures: $\chi_{\text{solute}} = 0.0347$

   Watch out

   A common mistake is to confuse molality (moles per kg of solvent) with molarity (moles per liter of solution). For molality, the denominator mass refers only to the solvent, which is why we can directly count 1000 g of water here.

   Tip

   For dilute aqueous solutions, you can use the quick approximation $\chi_{\text{solute}} \approx \frac{m}{55.56}$ where $m$ is the molality, since water contributes roughly 55.56 moles per kg. Here: $\frac{2.0}{55.56} \approx 0.036$, close to our answer.

   ✓Final answer

   The correct option is (C) $0.0347$.

2. CBSE 2026Set A1 markMCQ
   Q.34.2 g of sugar is present in 234.2 g of its aqueous solution. Then its molal concentration is

   (a) 0.1

   (b) 0.5

   (c) 5.5

   (d) 55.0

   ›Reveal solutionSolution

   Moles of sugar = 34.2/342 = 0.1; mass of solvent (water) = 234.2 - 34.2 = 200 g = 0.2 kg; molality = 0.1/0.2 = 0.5 m.

   Molar mass of sugar (sucrose) = 342 g/mol.

   Moles of sugar = 34.2/342 = 0.1 mol.

   Mass of solution = 234.2 g, mass of sugar = 34.2 g, so mass of water = 234.2 - 34.2 = 200 g = 0.2 kg.

   Molality = moles of solute / mass of solvent in kg = 0.1/0.2 = 0.5 mol/kg.

   ✓Final answer

   (b) 0.5.

3. CBSE 2026Set ANNUAL1 markMCQ
   Q.What will be the molarity of 30 ml of 0.5 M H2SO4 solution diluted to 50 ml?

   (a) 0.3 M

   (b) 0.03 M

   (c) 3 M

   (d) 0.13 M

   ›Reveal solutionSolution

   Dilution does not change the number of moles of solute, so M1V1 (before) = M2V2 (after).

   Given: M1 = 0.5 M, V1 = 30 mL, final volume V2 = 50 mL.

   Using the dilution law: M1V1 = M2V2

   0.5 x 30 = M2 x 50

   15 = 50 x M2

   M2 = 15/50 = 0.3 M

   ✓Final answer

   (a) 0.3 M.

4. CBSE 2026Set ANNUAL1 mark
   Q.Define mole fraction.

   ›Reveal solutionSolution

   Mole fraction expresses a component's amount relative to the total moles in a mixture, independent of temperature.

   For a solution containing components with n1, n2, n3, ... moles, the mole fraction of component 1 is defined as:

   x1 = n1 / (n1 + n2 + n3 + ...)

   It is a dimensionless quantity, and the mole fractions of all components in a solution always sum to 1. Unlike molarity, mole fraction does not depend on temperature, since it is based purely on the number of moles, not volume.

   ✓Final answer

   Mole fraction = (moles of a given component) / (total moles of all components in the solution).

5. CBSE 2026Set ANNUAL1 markMCQ
   Q.In an acid-base titrimetric analysis, the concentration of a sulphuric acid analyte is found to be 0.044 M. The strength of the acid in g/L is –

   (a) 0.44

   (b) 4.31

   (c) 2.15

   (d) 44.00

   ›Reveal solutionSolution

   Strength in g/L = molarity × molar mass = 0.044 × 98 ≈ 4.31 g/L, so option (B).

   The strength of a solution in grams per litre is related to its molarity by

   $\text{Strength (g/L)} = \text{Molarity (mol/L)} \times \text{Molar mass (g/mol)}.$

   The molar mass of sulphuric acid $H_2SO_4 = 2(1) + 32 + 4(16) = 98\ g\,mol^{-1}$.

   $\text{Strength} = 0.044 \times 98 = 4.312 \approx 4.31\ g\,L^{-1}.$

   ✓Final answer

   (B) 4.31 g/L.

6. CBSE 2025Set ANNUAL1 mark
   Q.Write the definition of molality.

   ›Reveal solutionSolution

   Molality expresses concentration as moles of solute per kilogram of SOLVENT (not solution), and unlike molarity it does not change with temperature.

   Definition:

   Molality (m) = (moles of solute) / (mass of solvent in kg)

   Unit: mol kg^-1 (also written 'molal', symbol m)

   Because molality is defined using the MASS of solvent (which does not change with temperature) rather than the VOLUME of solution (which expands/contracts with temperature), molality is a temperature-independent measure of concentration - unlike molarity, which does vary with temperature.

   ✓Final answer

   Molality = moles of solute per kilogram of solvent.

7. CBSE 2025Set ANNUAL1 markMCQ
   Q.What is the molarity of a solution with a mass of solute 10 kg mass and 100 litre volume?

   (a) 0.1 molar

   (b) 1 molar

   (c) 10 molar

   (d) 100 molar

   ›Reveal solutionSolution

   Molarity is defined as moles of solute per litre of solution; dividing the given amount of solute by the solution volume gives 0.1 M.

   Molarity is defined as:

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

   Note: as printed, the question states the solute quantity as '10 kg mass'; for the arithmetic to match any of the given options (0.1, 1, 10, 100 M) the intended quantity is 10 moles of solute (a common wording/printing slip in this recurring question, where 'kg' should read 'mol') - with no molar mass given, that is the only value that lets the problem be solved from the stated data. Taking the solute amount as 10 mol:

   $M = \dfrac{10\ \text{mol}}{100\ \text{L}} = 0.1\ \text{mol/L}$

   ✓Final answer

   (a) 0.1 molar.

8. CBSE 2025Set ANNUAL1 markMCQ
   Q.A solution contains 8 moles of solute and the mass of solvent is 4 kg. What is the molality of this solution?

   (a) 5 mol/kg

   (b) 8 mol/kg

   (c) 4 mol/kg

   (d) 2 mol/kg

   ›Reveal solutionSolution

   Molality is moles of solute per kilogram of solvent (not solution); here it works out to 8/4 = 2 mol/kg.

   Molality (m) is defined as:

   m = (moles of solute) / (mass of solvent in kg)

   Given:

   • moles of solute = 8 mol
   • mass of solvent = 4 kg

   m = 8 mol / 4 kg = 2 mol/kg

   Note the common trap here: molality uses the mass of the SOLVENT, not the total solution mass or volume (that would be molarity or a different quantity) — a frequent point of confusion tested in this exact question style.

   ✓Final answer

   (d) 2 mol/kg.

9. CBSE 2025Set ANNUAL1 mark
   Q.Define molality.

   ›Reveal solutionSolution

   Molality is defined as moles of solute per kilogram of solvent; unlike molarity, it does not depend on temperature since it is based on mass, not volume.

   Molality (denoted m) of a solution is defined as the number of moles of solute dissolved in one kilogram (1000 g) of solvent:

   Molality (m) = (number of moles of solute) / (mass of solvent in kg)

   Unit: mol kg^-1 (also written as 'm', e.g., a '1 molal' or '1 m' solution).

   Unlike molarity (which is based on the volume of the solution and therefore varies slightly with temperature, since volume expands/contracts with temperature), molality is based on the mass of the solvent, which does not change with temperature — so molality is a temperature-independent way of expressing concentration.

   ✓Final answer

   Molality is the number of moles of solute per kilogram of solvent: m = moles of solute / mass of solvent (in kg).

10. CBSE 2024Set A11 mark
    Q.The number of moles of solute present in one kilogram of the solvent is called \rule{2cm}{0.4pt}.

    ›Reveal solutionSolution

    Moles of solute per kilogram of solvent defines molality.

    Molality ($m$) is a concentration term that depends only on the mass of solvent (and is therefore temperature-independent):

    $m = \dfrac{\text{moles of solute}}{\text{mass of solvent in kg}} \;\; (\text{mol kg}^{-1})$

    Since the definition specifies moles of solute in one kilogram of solvent, the quantity described is molality.

    ✓Final answer

    molality

11. CBSE 2024Set ANNUAL1 markMCQ
    Q.Which of the following has no unit?

    (a) Molarity

    (b) Molality

    (c) Normality

    (d) Molar Fraction

    ›Reveal solutionSolution

    Mole fraction is a dimensionless ratio, so it has no unit.

    Molarity has units of $mol\,L^{-1}$, molality has units of $mol\,kg^{-1}$, and normality has units of $eq\,L^{-1}$. Mole (molar) fraction is defined as the ratio of moles of one component to the total moles of all components in the solution, $x_i = \dfrac{n_i}{\sum n}$ — since it is a ratio of like quantities, it is a pure number and carries no unit.

    ✓Final answer

    (iv) Molar Fraction

12. CBSE 2024Set ANNUAL1 mark
    Q.Write down the formula of Molarity.

    ›Reveal solutionSolution

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

    Molarity is one of the most widely used units of concentration. It is defined as the number of moles of solute dissolved in one litre (one cubic decimetre) of solution:

    $M = \frac{\text{moles of solute } (n)}{\text{volume of solution in litres } (V)}$

    Equivalently, in terms of mass, $M = \dfrac{w \times 1000}{M_{r} \times V(mL)}$, where $w$ is the mass of solute in grams and $M_r$ is its molar mass.

    ✓Final answer

    $M = \dfrac{n}{V(L)}\ mol\,L^{-1}$