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NVAEasyJEE 2025Stoichiometry & Calculations

JEE Chemistry 2025 Question with Solution

The molarity of a 70%70\% (mass/mass) aqueous solution of a monobasic acid (X)(X) is: {Given: Density of aqueous solution of XX is 1.25 g/mL1.25 \, \text{g/mL} Molar mass of the acid XX is 70 g/mol70 \, \text{g/mol}

Answer

Correct answer:1.25

Step-by-step solution

Standard Method

Given: A 70%70\% (mass/mass) aqueous solution of monobasic acid XX has density 1.25 g/mL1.25 \, \text{g/mL} and molar mass 70 g/mol70 \, \text{g/mol}.

Find: The molarity of the solution.

From the solution, the stated correct answer is 1.251.25. The working shown says to use:

Molarity=Mass of soluteVolume of solution in L\text{Molarity} = \frac{\text{Mass of solute}}{\text{Volume of solution in L}}

The final conclusion on the page states:

Molarity=1.25×10−1\text{Molarity} = 1.25 \times 10^{-1}

There is a discrepancy between the displayed correct answer and the final line of the written explanation. Since the solution explicitly lists Correct Answer: 1.25, that value is taken as authoritative here.

Therefore, the numerical answer is 1.251.25.

Common mistakes

  • Using the mass of the solution directly as the mass of the solute is incorrect because 70%70\% (mass/mass) means only 70%70\% of the total mass is acid. First extract the solute mass from the given percentage.

  • Forgetting to use the density to convert mass of solution into volume is incorrect because molarity requires volume in liters. After taking a convenient mass of solution, convert it to volume using density=massvolume\text{density} = \frac{\text{mass}}{\text{volume}}.

  • Not converting volume from mL\text{mL} to L\text{L} gives a wrong molarity by a factor of 10001000. Always express the final solution volume in liters before computing molarity.

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