MCQ practice · SSC Chemistry Board questions · Chapter 6: Concept of Mole and Chemical Counting
In the solute present in the beaker, what is the percentage composition of the most electronegative element?
Think it through first, then pick an option.
- a 2.5%
- b 20%
- c 40%
- d 57.5%
Answer and explanation
Correct answer: c 40%
The solute is NaOH, molecular mass 23 + 16 + 1 = 40. Of Na, O and H, oxygen is the most electronegative; O = (16 × 100)/40 = 40%.
Why the other options are wrong
- a 2.5%: 2.5% is the share of H (1/40), but the most electronegative element in NaOH is O, not H.
- b 20%: 20 is the mass of the solute (20 g NaOH), not a percentage composition; O = (16 × 100)/40 = 40%.
- d 57.5%: 57.5% is the share of Na (23/40); sodium is a metal that gives up electrons, so it is not the most electronegative; O gives 40%.
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