Logo
Jeeify
MCQEasyJEE 2024Concentration Terms (Molarity, Molality, etc.)

JEE Chemistry 2024 Question with Solution

The molality of a 0.8 M0.8 \, \text{M} H_2$$SO_4∗∗solutionwithdensity∗∗** solution with density **1.06 , \text{g/cm}^3$$ is:

  • A

    815×10−3 m815 \times 10^{-3} \, \text{m}

  • B

    0.815 m0.815 \, \text{m}

  • C

    1.5 m1.5 \, \text{m}

  • D

    0.5 m0.5 \, \text{m}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Molarity of H2_2SO4_4 is 0.8 M0.8 \, \text{M} and density of solution is 1.06 g/cm31.06 \, \text{g/cm}^3.

Find: Molality of the solution.

Take 1 L1 \, \text{L} of solution. Then moles of H2_2SO4_4 present are 0.8 mol0.8 \, \text{mol}.

Mass of 1 L1 \, \text{L} solution:

1.06 g/cm3×1000 cm3=1060 g1.06 \, \text{g/cm}^3 \times 1000 \, \text{cm}^3 = 1060 \, \text{g}

Mass of solute using molar mass 98 g/mol98 \, \text{g/mol}:

0.8×98=78.4 g0.8 \times 98 = 78.4 \, \text{g}

Mass of solvent:

1060−78.4=981.6 g=0.9816 kg1060 - 78.4 = 981.6 \, \text{g} = 0.9816 \, \text{kg}

Now apply the definition of molality:

m=moles of solutemass of solvent in kg=0.80.9816≈0.815 mm = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.8}{0.9816} \approx 0.815 \, \text{m}

This is equivalently written as:

0.815 m=815×10−3 m0.815 \, \text{m} = 815 \times 10^{-3} \, \text{m}

Therefore, the correct option is A.

Definition-Based Expansion

Given: Molarity is moles of solute per litre of solution, while molality is moles of solute per kilogram of solvent.

Find: Molality corresponding to the given 0.8 M0.8 \, \text{M} solution.

For 1 L1 \, \text{L} of solution, the number of moles of H2_2SO4_4 is 0.80.8.

Using density, total mass of this 1 L1 \, \text{L} solution is:

1.06 g/cm3×1000 cm3=1060 g1.06 \, \text{g/cm}^3 \times 1000 \, \text{cm}^3 = 1060 \, \text{g}

Mass of H2_2SO4_4 dissolved is:

0.8 mol×98 g/mol=78.4 g0.8 \, \text{mol} \times 98 \, \text{g/mol} = 78.4 \, \text{g}

So, mass of water present is:

1060−78.4=981.6 g1060 - 78.4 = 981.6 \, \text{g}

Convert to kilogram:

981.6 g=0.9816 kg981.6 \, \text{g} = 0.9816 \, \text{kg}

Hence,

m=0.80.9816=0.8149≈0.815 mm = \frac{0.8}{0.9816} = 0.8149 \approx 0.815 \, \text{m}

The solution also expresses this value as 815×10−3 m815 \times 10^{-3} \, \text{m}. Therefore, the correct option is A.

Common mistakes

  • Using molarity directly as molality is incorrect because molarity is based on volume of solution, whereas molality is based on mass of solvent. First convert the given data into kilograms of solvent.

  • Taking 1060 g1060 \, \text{g} as the mass of solvent is wrong because it is the mass of the entire solution. Subtract the mass of H2_2SO4_4 to get the mass of water.

  • Forgetting to convert grams to kilograms gives a value off by a factor of 10001000. Use 0.9816 kg0.9816 \, \text{kg}, not 981.6 g981.6 \, \text{g}, in the molality formula.

Practice more Concentration Terms (Molarity, Molality, etc.) questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Concentration Terms (Molarity, Molality, etc.)

Related questions