JEE Main · 2023 · Shift-IIeasyMOLE-106

A solution is prepared by adding 2\,g of X to 1 mole of water. Mass percent of X in solution is:

Some Basic Concepts (Mole Concept) · Class 11 · JEE Main Previous Year Question

Question

A solution is prepared by adding 2g2\,\text{g} of XX to 11 mole of water. Mass percent of XX in solution is:

Options
  1. a

    5%5\%

  2. b

    20%20\%

  3. c

    2%2\%

  4. d

    10%10\%

Correct Answerd

10%10\%

Detailed Solution

🧠 Mass percent uses the total solution mass 1mol1\,\text{mol} of water is 18g18\,\text{g}. Add the 2g2\,\text{g} of solute and you get 20g20\,\text{g} of solution.

🗺️ One quick step %mass of X=22+18×100=220×100=10%.\%\,\text{mass of }X = \frac{2}{2 + 18} \times 100 = \frac{2}{20} \times 100 = 10\%.

⚠️ The "divide by water" trap If you divide 22 by 1818 (water alone), you get 11.1%11.1\% — close to 10%10\% but wrong in concept. Always divide by total solution mass.

Answer: (d)\boxed{\text{Answer: (d)}}

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A solution is prepared by adding 2\,g of X to 1 mole of water. Mass percent of X in solution is: (JEE Main 2023) | Canvas Classes